ThE LIMIt OF A FUNCTION EVALUATING LIMITS – THE LIMIT LAWS Let's quickly reflect on what we covered in the previous section. We tried to evaluate limits using numerical and graphical methods, but we saw that these methods could only be used to make approximations. We also found that there are a number of pitfalls to guessing the values of limits, which usually leads to making wrong guesses. In this section, we will use foolproof methods to compute the EXACT values of limits. We will also examine ➔ how to combine the various laws to evaluate seemingly complicated limits, and ➔ Theorems that ease simplification of limits We start with the limit laws. The textbook I'm using highlights eleven limit laws. The first five deal with basic limit operations: addition, subtraction, multiplication, division and multiplication by a constant. The first five laws are stated below. Note that all eleven laws are valid if and only if the limits
lim f(x)
lim g(x)
and
x→a exist, and also if c is a constant:
x→a
Law 1 – Addition Law:
lim
x→a
[f(x)
+ g(x)]
lim f(x)
+
lim f(x)
–
=
x→a
lim g(x)
x→a
Law 2 – Difference Law:
lim
x→a
[f(x)
– g(x)]
=
x→a
lim g(x)
x→a
Law 3 – Constant Multiple Law:
lim
x→a
[c f(x)]
=
c lim f(x) x→a
Law 4 – Product Law:
lim
x→a
[f(x) g(x)]
lim f(x)
=
x→a
×
lim g(x)
x→a
Law 5 – Quotient/Division Law:
lim
x→a
f(x) g(x)
=
lim f(x)
x→a
lim g(x)
x→a
[On the condition that the denominator IS NOT ZERO]
The sixth law is the power law, which is derived from the product law, by putting g(x) = f(x) and using the product law repeatedly:
lim
[f(x)]n
x→a
lim f(x)
=
n
x→a
Law 6 – Power Law: Laws 7 and 8 are critically important, if we are to apply any of the limit laws:
lim c = c
Law 7
lim x = a
Law 8
x→a
x→a
Law 9 is obtained from Law 6 by putting f(x) = x and applying Law 8. From Law 6,
lim
x→a
[f(x)]n
lim f(x)
=
n
x→a
If we put f(x) = x, we have
lim [x]n
lim x
=
x→a
n
x→a
Applying Law 8 gives
lim
x→a
xn =
an Law 9
The next law is the Root Law:
lim
x→a
n
n
√f(x)
=
√
lim f (x)
x→a
Where n is a positive integer. We use this law with the assumption that the limit is greater than zero, if n is an even number.
Law 10
Finally, we have the eleventh law, which is basically a slight modification of law 10. Specifically, this law is obtained from law 10 by putting f(x) = x and applying law 8. From Law 10, we know that
lim
x→a
n
n
√f(x)
Putting f(x) = x gives
=
√ lim
x→a
f (x)
n
lim
x→a
n
√x
√ lim
=
x→a
x
From Law 8,
lim x = a
x→a Therefore,
lim
x→a
n
n
√x
=
√a
Law 11 For the rest of this section, we'll use these 11 laws in computing the exact values of limits. In applying any of these 11 limit laws, there's one general rule that applies to all. If f is a polynomial/rational function, and a is in the domain of f, then,
lim
x→a
f(x) = f(a)
Law 12 Functions that satisfy Law 12 are said to be continuous. Thus, their limits can be evaluated by using direct substitution. Bear in mind that there are exceptions, as we'll see later in this section.
Example 1 Estimate the value of the limit
lim (5x2 – 2x + 3)
x→4 (a)
by evaluating g(x) = 5x2 – 2x + 3 for values of x that approach 4.
(b)
from a graph of g.
(c)
by using the appropriate limit laws. Justify each step.
Solution x
g(x)
x
g(x)
3.5
57.25
4.5
95.25
3.7
64.05
4.3
86.85
3.9
71.25
4.1
78.85
3.99
74.62
4.01
75.38
3.995
74.81
4.001
75.04
3.999
74.96
4.0005
75.01
3.9995
74.98
4.0001
75.00
3.9999
74.99
4.00001
75.00
The question asks that we use numerical evidence to guess the value of the limit. As usual, we draw up a table of values for g(x) using values of x that approach 4 from both sides. The table clearly shows that g(x) approaches 75 as x gets closer and closer to 4 from either side. Thus, using this table, we assume that
lim
x→4
(5x2 – 2x + 3)
=
75
Therefore, numerical evidence shows that the value of the limit is 75.
Next, we'll use graphical evidence to guess the value of the limit. If we graph the function g(x) = 5x2 – 2x + 3 in the viewing rectangle [–6, 6] by [–4, 85], we get this:
g(x) → 75 as x → 4 from both sides
From the graph, it is clear that g(x) → 75 as x → 4 from the left. Thus,
lim
X→4–
(5x2 – 2x + 3)
=
75
Likewise, g(x) → 75 as x → 4 from the right. So,
lim
X→4+
(5x2 – 2x + 3)
=
75
Since the left and right hand limits exist and are equal, it is safe to guess that
lim
x→4
(5x2 – 2x + 3)
=
75
OK, Enough with the guessing. Let's do some real math. This time, we'll apply the limit laws; no graphs, no tables. Here goes: Using the addition and subtraction laws (laws 1 and 2),
lim (5x2 – 2x + 3)
x→4
lim 5x2
=
–
x→4
lim 2x
x→4
+
lim
x→4
(1)
3
You can see that this step is pretty much like expanding an algebraic expression. Next, we deal with the three limits on the RHS of (1) above. Now study the following steps carefully, as we combine the limit laws:
lim 5x2
x→4
–
lim 2x
x→4
+
lim
x→4
3
=
5 lim x2 x→4
–
2 lim
x→4
x
+
lim
x→4
3
Law 3
Now that the limits have been reduced to their simplest possible forms, the next step is to evaluate them via direct substitution:
5 lim x2 x→4
–
2 lim
x→4
x
lim
+
x→4
3
=
5(42) – 2(4) + 3
=
80 – 8 + 3
=
75
At this point, we can categorically say that
lim
x→4
(5x2 – 2x + 3)
=
75
Example 2 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
lim (x3 + 2)(x2 – 5x)
x→3
Solution From the question, we see that
f(x) = (x3 + 2)(x2 – 5x) There are two ways of evaluating this limit: ●
Since the function is a product, we can start by breaking the limit into a product of two limits. From the product law,
lim (x3 + 2)(x2 – 5x)
x→3
lim (x3 + 2)
=
lim (x2 – 5x)
×
x→3
x→3
On the other hand, ●
We can simply expand f, and evaluate:
lim (x3 + 2)(x2 – 5x)
x→3
lim
=
x→3
(x
– 5x4 + 2x2 – 10x)
5
Either way, we get the same answer. Let's use the first method:
lim (x3 + 2)(x2 – 5x)
x→3
=
lim (x3 + 2)
=
lim
x3
x→3
= =
[(3)
3
]
+2
[27 + 2]
lim (x2 – 5x)
×
x→3
lim 2
+
[(3)
2
×
[9
x→3
]
– 5(3)
]
– 15
Try using the second method. Here's a hint: use laws 1, 2, 3, 8, and 9. Bear in mind, that your answer MUST BE –174.
lim
×
x→3
×
(Law 4)
x→3
x2
–
5 lim
x→3
(Laws 7, 8, 9) =
– 174
x
(Laws 1, 2, 3)
Note: When you're dealing with polynomials, you can always verify your answer by evaluating f at a; that is, by computing f(a). This is called direct substitution. It also works for rational functions on ONE CONDITION: THAT THE DENOMINATYOR DOES NOT EVALUATE TO ZERO).
Example 3 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
2
(x4 + x2 – 6)
lim
(x4 + 2x + 3)
x→1
Solution Generally, understanding the kind of function you're dealing with helps you decide which limit law to use first. In this case, it is obvious that we're dealing with a quotient function. Hence, we apply the quotient law first:
lim
x→1
2
2
lim (x4 + x2 – 6)
2
(x + x – 6) 4
x→1
=
(x4 + 2x + 3)
lim (x4 + 2x + 3)
x→1 Then, we use the power law:
2
lim (x4 + x2 – 6)
lim (x4 + x2 – 6)
x→1
x→1
=
lim (x4 + 2x + 3)
2
x→1
lim (x4 + 2x + 3)
2
x→1 Using Laws 1, 2 and 3, we have
lim (x4 + x2 – 6)
2
lim x4
x→1
x→1
lim x2
+
–
x→1
=
lim (x4 + 2x + 3)
lim 6
x→1
2
x→1
lim x4
x→1
+
2 lim
x→1
x
+
2
lim 3
2
x→1
Using direct substitution, and laws 7, 8, and 9, we have
lim x4
x→1
lim x
4
x→1
+
+
lim x2
–
x→1
2 lim
x→1
x
+
2
lim 6
lim 3
= 2
x→1
(x4 + x2 – 6) (x + 2x + 3) 4
+ (1)2 – 6)2
((1)
+ 2(1) + 3)2
4
x→1
Therefore,
lim
((1)
4
x→1
2 =
4 9
(–4)2 =
= (6)2
4 9
Example 4 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
lim
(t + 1)9 (t2 – 1)
t→ –2
Solution Here, the function is f(t) = (t + 1)9 (t2 – 1). One way of solving this limit is by expanding f, but the drawback is that f will be too long when expanded, and let's not forget that that there is a slight chance of making a mistake during the expansion, An alternative method is to split the limit into two, since f is a product (like example 2):
lim
=
(t + 1)9 (t2 – 1)
t→ –2
lim (t + 1)9
t→ –2
lim (t2 – 1)
×
(Product Law)
t→ –2
Then,
lim (t + 1)9
t→ –2
lim (t2 – 1)
×
t→ –2
lim t
=
lim 1
+
t→ –2
lim (t + 1)
=
9
t→ –2
t→ –2
–
lim (t2 – 1)
×
t→ –2
lim t2
×
9
(Power Law)
t→ –2
lim 1
(Addition & Subtraction Laws)
t→ –2
Using Laws 7, 8 and 9, we have
lim t
t→ –2
lim 1
+
9
t→ –2
lim t2
×
–
t→ –2
lim 1
t→ –2
=
[(–2)
=
–3
]9 × [(–2)
+ 1
2
Therefore,
lim
(t + 1)9 (t2 – 1)
t→ –2
=
–3
Example 5 Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
lim
u→ –2
√u
4
+ 3u + 6
Solution Since f(u) = u4 + 3u + 6, then Law 10 (the Root Law) gives
lim
u→ –2
√u
4
+ 3u + 6
√ lim
=
u→ –2
[u
4
+ 3u + 6]
We then use the addition law, together with the constant multiple law to simplify f:
√ lim
u→ –2
[u
4
+ 3u + 6]
=
√
lim
u→ –2
(u ) 4
+
lim
u→ –2
(3u)
+
lim
u→ –2
(6)
]
– 1
Evaluating the limits gives
√
lim
u→ –2
(u ) 4
+
lim
u→ –2
(3u)
lim
+
(6)
u→ –2
=
√(–2)
=
√16
4
+ 3(–2) + 6
– 6 + 6
=
4
Hence,
lim
u→ –2
√u
4
=
+ 3u + 6
4
Example 6 (a)
What is wrong with the equation?
x2 + x – 6
=
x – 2 (b)
x + 3
In view of part (a), explain why the equation
x2 + x – 6
lim
x→2
lim (x + 3)
=
x – 2
x→2
is correct.
Solution Clearly, there is nothing wrong with the equation
x2 + x – 6
=
x – 2
x + 3
because, if we evaluate the expression on the left, we have
x2 + x – 6
=
x – 2
(x + 3)(x – 2)
=
x – 2
x + 3
(i)
Hence, the equation is correct. For this very reason, the equation
lim
x→2
x2 + x – 6 x – 2
=
lim (x + 3)
(ii)
x→2
is also correct. Bear in mind that,although (ii) is correct, the limit on the left, that is
lim
x→2
x2 + x – 6 x – 2
DOES NOT EXIST. This is because the denominator would evaluate to
zero using either the limit laws or direct
substitution. But if the function is simplified, we get the limit on the right side of (ii) as shown from (i). We can prove that (ii) is correct in two ways: ➔ By simplifying the function itself:
lim
x→2
x2 + x – 6 x – 2
=
lim
x→2
(x + 3)(x – 2) (x – 2)
=
lim (x + 3)
x→2
➔ OR by using the limit laws to simplify the limit entirely:
lim (x2 + x – 6)
x2 + x – 6
lim
x→2
x→2
=
x – 2
(Quotient Law)
lim
(x – 2)
x→2 Simplifying the numerator gives
lim
lim (x + 3)(x – 2)
x→2
lim
(x + 3)
x→2
x→2
(x – 2)
=
lim
x→2
(Product Law)
(x – 2)
lim
x→2
(x – 2)
Canceling out equal limits results in
lim
x→2
(x + 3)
lim
x→2
(x – 2) =
lim
x→2
(x – 2)
lim (x + 3)
x→2
Either way, the equation is correct.
Example 7 Evaluate the limit, if it exists.
lim
x→ –3
x2 – x + 12 x +3
Solution To start with, we use direct substitution; that is, we put x = –3. If we do this, we find that the denominator equals zero. To this end, we conclude that the limit does not exist. Then again, there's a possibility we are jumping to conclusions here. The truth is, we are. Here's one rule you should STRICTLY adhere to when evaluating the limit of a rational function: TRY TO SIMPLIFY THE NUMERATOR OR DENOMINATOR WHEREVER POSSIBLE. In some cases, it is possible to simplify both numerator and denominator, so that you'll end up canceling out equal values. This is what we did in example 6. In this case, the function CANNOT be simplified any further. Hence, we can officially say that the limit DOES NOT EXIST. For cases like this, one way of verifying your result is by graphing the function. Since a limit does not exist when x = –3, you should expect to see some kind of discontinuity. If you look at the graph carefully, you'll agree that the limit is actually infinite (both positive and negative). Thus, the actual answer should be
lim
x→ –3
x2 – x + 12 x +3
=
± ∞
Example 8 Evaluate the limit, if it exists.
x + 2
lim
x→ –2
x – x – 6 2
Solution We start by simplifying the denominator:
x + 2
lim
x→ –2
=
x – x – 6 2
x + 2
lim
x→ –2
(x + 2)(x – 3)
1
lim
=
x→ –2
(x – 3)
Using direct substitution,
1
lim
x→ –2
=
(x – 3)
1
lim
x→ –2
(– 2 – 3)
lim
=
x→ –2
1
1
=
–5
–5
On the other hand, we can use the limit laws:
x→ –2
lim
1
lim
=
(x – 3)
x→ –2
lim
1
x→ –2
=
lim (x – 3)
lim
x→ –2
x→ –2
which equals
lim
x→ –2
1
=
(– 2 – 3)
lim
x→ –2
1 –5
Thus,
x + 2
lim
x→ –2
x – x – 6 2
=
1 –5
=
1 –5
x
–
1
lim
x→ –2
3
Example 9 Evaluate the limit, if it exists.
x2 + x – 2
lim
x→1
x2 – 3x + 2
Solution Here, we see that direct substitution is not an option, for obvious reasons. Neither can we use the quotient law at this stage. Therefore, we can try to evaluate the limit, by simplifying it first. In this case, we see that both numerator and denominator can be factorized:
x2 + x – 2
lim
x→1
x – 3x + 2
(x + 2)(x – 1)
lim
=
2
x→1
(x – 2)(x – 1)
=
lim
x→1
x +2 x – 2
Now we can use the limit laws:
lim (x + 2)
x +2
lim
x→1
x→1
=
x – 2
x→1
lim
lim (x + 2)
x→1
lim (x – 2)
(Quotient Law)
lim (x – 2)
x→1
=
lim
x→1
x→1
x x
+ –
lim
x→1
lim
x→1
2 (Addition & Subtraction Laws) 2
Then, using laws 7 and 8, we have
lim
x→1
lim
x→1
x x
+ –
lim
x→1
lim
x→1
2 1 + 2
=
1 – 2
2
=
–3
Therefore,
x2 + x – 2
lim
x→1
x2 – 3x + 2
=
–3
Example 10 Evaluate the limit, if it exists.
lim
h→0
(h – 5)2 – 25 h
Solution Like example 9, using direct substitution or the quotient rule at this stage is not a good idea. The function needs to be simplified first. We start by expanding the numerator:
(h – 5)2 – 25
lim
h→0
=
h
h2 – 10h
lim
h→0
h→0
h(h – 10)
h→0
h→0
h
lim (h – 10)
=
h
h2 – 10h + 25 – 25
lim
=
h
lim
=
h
(h2 – 10h + 25) – 25
lim
=
h→0
You can see that we have reduced
(h – 5)2 – 25
lim
h→0
lim (h – 10)
to
h
h→0
Now we can use the limit laws to evaluate the limit
lim (h – 10)
lim h
=
h→0
–
h→0
=
0 – 10
=
– 10
lim 10
(Subtraction Law)
h→0
(Laws 7 and 8)
Therefore,
(h – 5)2 – 25
lim
h→0
=
h
– 10
Example 11 Evaluate the limit, if it exists.
x3 – 1
lim
x→1
x2 – 1
Solution We need to simplify both numerator and denominator of the function:
lim
x3 – 1
x→1
lim
=
x –1 2
x→1
(x – 1)(x2 + x + 1)
=
(x – 1)(x + 1)
lim
x→1
x2 + x + 1 x+1
Using direct substitution,
x2 + x + 1
lim
x→1
=
x+1
(1)2 + (1) + 1 (1) + 1
=
3 2
Therefore,
lim
x→1
x3 – 1 x –1 2
=
3 2
Here, I used direct substitution to evaluate the limit, after reducing the function to a simpler form. Now, try using the limit laws to evaluate the limit. One thing I've noticed is that, anytime you evaluate a limit, you'll always end up using laws 7, 8 and/or 9 (usually in the final stage).
Example 12 Evaluate the limit, if it exists.
(1 + h)4 – 1
lim
h→0
h
Solution Expanding the numerator gives
lim
(1 + h)4 – 1
h→0
h
lim
=
lim
(1 + 4h + 6h2 + 4h3 + h4) – 1
h→0
h
4h + 6h2 + 4h3 + h4 – 1 + 1
h→0
=
lim
=
h h(4 + 6h + 4h2 + h3)
h→0
h
4h + 6h2 + 4h3 + h4
lim
=
h→0
h
lim (4 + 6h + 4h2 + h3)
=
h→0
Using direct substitution,
lim (4 + 6h + 4h2 + h3)
=
h→0
4 + 6(0) + 4(0)2 + (0)3
=
4
Again, I have used direct substitution to evaluate the limit after reducing it to a simpler form (just like example 11). Here's another task: use the limit laws to evaluate the limit. Justify each step by indicating the appropriate limit laws you applied.
Example 13 Evaluate the limit, if it exists.
lim
t→9
9 – t 3 – √t
Solution There are two possible approaches to solving this limit: ➔ Expressing the numerator as a difference of squares, and canceling out the common factors:
lim
t→9
9 – t 3 – √t
=
2 2 lim (3) – (√t)
t→9
=
3 – √t
(3 – √t)(3 + √t)
lim
t→9
3 – √t
=
lim (3 + √t) t→9
OR ➔ Rationalize the denominator:
lim
t→9
=
9 – t 3 – √t
lim
t→9
=
lim
t→9
(9 – t)(3 + √t) (9 – t)
9 – t 3 – √t =
×
3 + √t 3 + √t
lim (3 + √t)
t→9
=
lim
t→9
(9 – t)(3 + √t) (3 – √t )(3 + √t)
Either way, we end up with one limit:
lim (3 + √t)
t→9
which can easily be evaluated. Using the limit laws:
lim (3 + √t)
t→9
=
lim 3
+
=
lim 3
+
=
3 + √9
t→9
t→9
=
lim √t
(Addition Law)
t→9
√lim
t→9
t
(Root Law) (Laws 7 and 8)
6
Therefore,
9 – t
lim
t→9
lim (3 + √t)
=
3 – √t
=
t→9
6
Example 14 Evaluate the limit, if it exists.
lim
t→2
t2 + t – 6 t2 – 4
Solution Understand this: when evaluating the limit of a rational function, there is often the need to simplify it. The aim of simplifying the limit is to reduce it to a relatively simpler form, or perhaps its simplest possible form so that the appropriate limit laws can be applied. Here, we factorize both numerator and denominator:
lim
t2 + t – 6
t→2
t – 4
(t + 3)(t – 2)
lim
=
2
t→2
(t + 2)(t – 2)
=
lim t→2
(t + 3) (t + 2)
At this point, it will do absolutely no harm to perform direct substitution:
lim t→2
(t + 3) (t + 2)
2+3
=
2+2
=
5 4
Try using the limit laws to evaluate the limit.
Example 15 Evaluate the limit, if it exists.
lim
t→0
√2 – t – √2 t
Solution We cannot use direct substitution, nor the quotient law, for obvious reasons. Neither can we simplify the numerator by factorization. So, how do we simplify this limit?
When attempting to evaluate the limit of a rational function whose numerator or denominator (or perhaps both) is irrational, one good approach is rationalizing the irrational expression first. This process should simplify the function considerably. This is demonstrated in example 13. In this case, we rationalize the numerator:
√2 – t
lim
t→0
– √2
√2 – t
+ √2 √ 2 – t + √2
×
t
which gives
(2 – t) + (√2 × √2 – t) –
lim
t→0
(√2 × √2 – t) – 2
(2 – t) + (√2 (2 – t)) –
lim
=
t→0
t(√ 2 – t + √2 )
t(√ 2 – t + √2 )
After simplifying the numerator, we end up with 2 – t – 2
lim
t→0
t(√ 2 – t + √2 )
–t
lim
=
t→0
=
t(√ 2 – t + √2 )
–1
lim
t→0
(√ 2 – t + √2 )
Now that the limit is simplified considerably, we can use the limit laws to evaluate it. Thus,
lim –1 –1
lim
t→0
t→0
=
(√ 2 – t + √2 )
(Quotient Law)
lim (√ 2 – t + √2 ) t→0
lim –1 t→0
=
lim √ 2 – t t→0
(Addition Law) +
lim √ 2 t→0
lim –1 =
t→0
√
lim (2 – t) + t→0
√
(Root Law)
lim 2 t→0
–1
=
(Laws 7 & 8)
√ 2 – 0 + √2 –1
=
√2
–1
=
+ √2
2√2
If you choose to rationalize the answer, then you'd have –1
×
2√2
2√2
=
2√2
– 2√2
8
=
– √2
4
Thus,
lim
t→0
√2 – t – √2 t
=
lim
t→0
–1
(√ 2 – t + √2 )
=
(√2 (2 – t)) – 2
–1
2√2
=
– √2
4
Example 16 Evaluate the limit, if it exists.
x4 – 16 x – 2
lim
x→2
Solution As usual, we simplify the function. We start by expressing the numerator as a difference of squares:
x4 – 16 x – 2
lim
x→2
(x2)2 – (4)2 x – 2
lim
=
x→2
=
(x2 – 4)(x2 + 4) x – 2
lim
x→2
We split the numerator even further to get
lim
x→2
(x2 – 4)(x2 + 4) x – 2
(x – 2)(x + 2)(x2 + 4) x – 2
lim
=
x→2
which leaves us with
lim (x + 2)(x2 + 4)
lim (x3 + 4x + 2x2 + 8)
=
x→2
x→2
Using direct substitution,
lim (x3 + 4x + 2x2 + 8)
x→2
=
(2)3 + 4(2) + 2(2)2 + 8
=
32
Alternatively, you can use the limit laws.
Example 17 Evaluate the limit, if it exists.
lim
x→1
1 x –1
–
2 x2 – 1
Solution At a glance, it would seem that the subtraction law would be the best option. On the contrary, it would be the worst:
lim
x→1
1 x –1
–
2 x2 – 1
=
lim
x→1
1 x –1
–
lim
x→1
2 x2 – 1
Clearly, it won't work because both denominators will evaluate to zero, and we know what that means! A better option is to evaluate the expression in the bracket first:
x→1
1 x –1
=
lim
lim
x→1
–
2 x –1 2
x2 – 2x + 1 (x – 1)(x – 1)(x + 1)
=
2 lim (x – 1) – 2(x – 1)
x→1
=
(x – 1)(x2 – 1)
lim
x→1
=
(x – 1)(x – 1) (x – 1)(x – 1)(x + 1)
x2 – 1 – 2x + 2 (x – 1)(x – 1)(x + 1)
lim
x→1
=
lim
x→1
1 x+1
So, by simply evaluating the bracketed expression, we have reduced
1 x –1
lim
x→1
2 x2 – 1
–
lim
to
x→1
1 x+1
This further proves the importance of simplifying a function before attempting to evaluate its limit. It makes the task a whole lot easier. Using direct substitution,
lim
x→1
1 x+1
1 1+1
=
1 2
=
Thus,
1 x –1
lim
x→1
2 x2 – 1
–
1 x+1
lim
=
x→1
=
1 2
Using direct substitution, the limit evaluates to ½. Now try using the limit laws to evaluate the limit.
Example 18 Evaluate the limit, if it exists.
lim
h→0
(3 + h)–1 – 3–1 h
Solution We simplify the numerator:
lim
h→0
=
(3 + h)–1 – 3–1 h
lim
3– 3–h 9 + 3h
h→0
=
lim
=
1 3+h
lim
h→0
–h 9 + 3h
h→0
h
1 3
lim
=
h
=
3 – (3 + h) 3(3 + h)
h→0
h
lim
=
–
lim
h→0
–h (9 + 3h)
h
×
1 h
–1
h→0 (9 + 3h)
Using the limit laws,
lim – 1
lim
–1
h→0 (9 + 3h)
h→0
=
(Quotient Law)
lim (9 + 3h)
h→0
lim – 1
h→0
=
lim 9
h→0
+
(Addition & Constant Multiple Laws) 3 lim h h→0
Using laws 7 and 8, we have
–1
–1
=
9 + 3(0)
9
Therefore,
(3 + h)–1 – 3–1 h
lim
h→0
=
–1 9
Example 19 Evaluate the limit, if it exists.
1 x
lim
x→2
1 2
– x – 2
Solution First, we simplify the function:
lim
1 x
1 2
–
x→2
=
x – 2
2 – x 2x
lim
x→2
x – 2
lim
=
=
x→2
lim
x→2
2 – x 2x
×
1 x – 2
2 – x (x – 2)(2x)
Next, we multiply the denominator by a minus sign. This allows us to cancel out the common factors:
lim
x→2
2 – x
lim
=
– [(x – 2)(2x)]
x→2
2 – x (–x + 2)(–2x)
=
lim
x→2
2 – x (2 – x)(–2x)
=
Using limit laws,
lim 1
1
lim
x→2
x→2
=
–2x
(Quotient & Constant Multiple Laws)
–2 lim x x→2
=
1 –2(2)
=
1 –4
(Laws 7 and 8)
Thus,
lim
x→2
1 x
– x – 2
1 2
=
1 –4
In the next set of examples (20 – 22), we learn how to evaluate limits of absolute functions.
lim
x→2
1 –2x
Example 20 Evaluate the limit, if it exists.
lim
|x + 4| x + 4
x→–4 –
Solution From the definition of an absolute function,
|x|
=
x
if x ≥ 0
–x
if x < 0
Let's assume
f(x)
|x + 4| x + 4
=
Therefore,
|x + 4| x + 4
x + 4 x + 4
if x ≥ 0
– (x + 4) x + 4
if x < 0
=
which equals
|x + 4| x + 4
=
1
if x ≥ 0
–1
if x < 0
In other words, ➔ f(x) = 1 for x ≥ 0
(i)
➔ f(x) = –1 for x < 0
(ii)
The task here is to evaluate the limit of f as x → –4 from the left. This means that we consider case (ii) only. Thus, since f(x) = –1 for x < 0, then
lim f(x) = –1
x→–4 –
You can verify the answer by graphing f.
Example 21 Evaluate the limit, if it exists.
lim
x→1.5
2x2 – 3x |2x – 3|
(direct substitution; Law 7)
Solution First, we simplify the function (if possible):
lim
x→1.5
2x2 – 3x |2x – 3|
lim
=
x→1.5
x(2x – 3) |2x – 3|
From the definition of an absolute function,
g(x)
=
x(2x – 3) |2x – 3|
x(2x – 3) 2x – 3
if x ≥ 0
x(2x – 3) – (2x – 3)
if x < 0
=
which equals
g(x)
x(2x – 3)
=
=
|2x – 3|
x
if x ≥ 0
–x
if x < 0
This means that ➔ g(x) = x for x ≥ 0
(i)
➔ g(x) = – x for x < 0
(ii)
So, ➔ For case (i).
lim
g(x) = 1.5
(Law 8)
lim
g(x) = –1.5
(Law 8)
lim
g(x)
x→1.5 + ➔ For case (ii). x→1.5 – Clearly, x→1.5 +
≠
lim
x→1.5 –
g(x)
In other words, we are saying that
lim
2x2 – 3x x→1.5 |2x – 3|
DOES NOT EXIST
Example 22 Evaluate the limit, if it exists.
lim
x→0
1 x
–
1 |x|
Solution Following the definition of an absolute function,
h(x)
1 x
=
1 |x|
–
1 x
–
1 x
1 x
–
–
if x ≥ 0
= 1 x
if x < 0
This becomes
h(x)
=
0
if x ≥ 0
2 x
if x < 0
So, to evaluate
lim
x→0
1 x
1 |x|
–
we consider two cases (just as we did in example 21 above): ➔ h(x) = x for x ≥ 0
(i)
➔ h(x) = 2/x for x < 0
(ii)
Now, since h(x) = x for x ≥ 0 (case (i)), then
lim h(x) = 0
(Law 7)
x→0 +
On the other hand, there's case (ii), which says that h(x) = 2/x for x < 0. Therefore, by direct substitution,
lim h(x) –
2 0
=
x→0
This means a limit DOES NOT EXIST for case (ii). In summary,
lim h(x) = 0
and
x→0 +
lim h(x)
x→0 –
Since these one sided limits are not equal, it follows that
lim
x→0
1 x
–
1 |x|
Example 23 Let
F(x)
=
Determine if
lim F(x) x→1
exists, and sketch the graph of F.
x2 – 1 |x – 1|
DOES NOT EXIST
DOES NOT EXIST
Solution Below is a graph of F:
To evaluate the limit, the first step to be taken is to rewrite F using the definition of the absolute function:
F(x)
=
(x + 1)(x – 1) x–1
x2 – 1
if x ≥ 0
=
|x – 1|
(x + 1)(x – 1) –(x – 1)
if x < 0
Thus,
F(x)
=
F(x)
=
OR
(x + 1)
if x ≥ 0
–(x + 1)
if x < 0
x+1
if x ≥ 0 if x < 0
–x – 1
Now that F has been broken down into a simplified piecewise defined function, we can evaluate the limit by considering two simple cases: ➔ F(x) = x + 1 for x ≥ 0
(i)
➔ F(x) = – x – 1 for x < 0
(ii)
CASE 1: F(x) = x + 1 for x ≥ 0 Thus,
lim F(x)
=
x→1 +
1 + 1
=
2
(Direct Substitution)
=
–2
(Direct Substitution)
CASE 2: F(x) = – x – 1 for x < 0 Thus,
lim F(x)
=
x→1 –
–1 – 1
At this point, it is obvious that
lim F(x)
lim F(x)
≠
x→1 +
x→1 –
This means that
lim F(x)
DOES NOT EXIST
x→1
The graph of F above clearly verifies this result.
Example 24 The signum (or sign) function, denoted by sgn, is defined by
sgn x
=
–1
if x < 0
0
if x = 0
1
if x > 0
(a)
Sketch the graph of this function.
(b)
Find each of the following limits or explain why it does not exist.
(i)
lim sgn x x→0
(ii)
lim |sgn x| x→0
Solution Based on the definition of the signum function, we have this graph:
The graph is quite straightforward, and will be used to evaluate the limits in question:
(i)
lim sgn x x→0
To evaluate this limit, we need to approach it from both sides, that is, find the left and right-hand limits. From the graph, we see that
lim sgn x
=
x→0 +
1
and
lim sgn x
x→0 –
=
–1
Since the two one-sided limits are not equal, it follows that
lim sgn x
DOES NOT EXIST
x→0
(ii)
lim |sgn x| x→0
This one is a little tricky, but recall the definition of an absolute function:
|x|
=
x
if x ≥ 0
–x
if x < 0
Using this definition, then
|sgn x|
–(–1)
if x < 0
0
if x = 0
1
if x > 0
=
which results in
|sgn x|
=
Graphically, we have
1
if x < 0
0
if x = 0
1
if x > 0
From the graph,
lim |sgn x|
=
x→0 +
1
lim |sgn x|
and
=
x→0 –
1
which means
lim |sgn x|
=
1
x→0 Again, always remember that, when dealing with absolute functions, always apply this rule:
|x|
=
x
if x ≥ 0
–x
if x < 0
On my graphing calculator (CASIO Fx-9750G PLUS), the absolute function |x| is input as abs x. The abs function is one of the many built in functions of a typical graphing calculator. Consult your calculator's manual for more info on how to use this function, both for calculations and graphing.
Example 25 Let
h(x)
=
x
if x < 0
x
if x = 0
8–x
if x > 0
2
Sketch the graph of h, and evaluate each of the following limits, if it exists:
(i)
lim h(x)
x→0 +
(ii)
lim h(x) x→0
(iii) lim
x→1
h(x)
(iv)
lim h(x) x→2
Solution This is another piecewise defined function, defined by three individual functions on three different domains:
CASE 1:
h(x) = x
if x < 0
➔ CASE 2:
h(x) = x2
if 0 ≤ x ≤ 2
➔ CASE 3:
h(x) = 8 – x
➔
if x > 2
The values of the given limits depends solely on at least one of these cases.
(i)
lim h(x)
x→0 +
Here, we want to find the value of the limit for values of x that are greater than zero. This means we consider case 2, as it deals only with values of x greater than zero but less than/equal to 2. Thus, by direct substitution,
lim h(x)
x→0 +
(ii)
=
lim (x2)
x→0 +
=
02
lim h(x) x→0
To solve this limit, we condense it into two simpler limits:
lim h(x)
(This limit points to CASE 1)
lim h(x)
(This limit points to CASE 2)
x→0 –
x→0 +
=
0
From CASE 1, h(x) = x if x < 0. Therefore,
lim h(x)
x→0 –
lim
=
x→0 –
x
=
0
For CASE 2, h(x) = x2 if 0 ≤ x ≤ 2. Thus,
lim h(x)
x→0 +
lim (x2)
=
x→0 +
=
02
=
0
Since, both cases yield the same result, we conclude that
lim h(x)
x→0 –
(iii) lim
x→1
=
0
h(x)
Here, we are to evaluate the limit of h as x approaches 1. Clearly, this would fit into CASE 2, as described by its domain, 0 ≤ x ≤ 2. Besides, don't forget that h(x) = x2 for 0 ≤ x ≤ 2. Therefore, by direct substitution,
lim h(x)
lim (x2)
=
x→1
=
x→1
12
=
1
Therefore,
lim h(x)
=
x→1
(iv) lim
x→2
1
h(x)
To solve this limit, we consider two cases:
lim h(x)
(which fits into CASE 2)
lim h(x)
(which fits into CASE 3)
x→2 – and
x→2 +
From CASE 2, h(x) = x2 for 0 ≤ x ≤ 2. Therefore,
lim h(x)
x→2 –
lim (x2)
=
x→2 –
=
22
=
4
and, from CASE 3, h(x) = 8 – x for x > 2. Therefore,
lim h(x)
x→2 + In summary,
lim h(x)
x→2 –
lim (8 – x)
=
=
=
x→2 +
4
and
8–2
lim h(x)
x→2 +
=
=
Clearly, this is an indication that
lim h(x) x→2
DOES NOT EXIST
In the following example, we take a look at a special function.
6
6
Example 26 In this example, we'll look at the greatest integer function. By definition, the greatest integer function, written as y = [x] , is the largest integer that is less than or equal to x . For example, [3] = 3, [3.6] = 3, [4.7] = 4, [π] = 3, and so on. The greatest integer function is graphed below:
(A) Using this definition, evaluate
(i)
lim [x]
x→ –2 +
From the graph, if we trace the lines, we see that [x] → –2 as x → –2 from the right. Therefore,
lim [x]
x→ –2 +
(ii)
=
–2
lim [x]
x→ –2
To solve this limit, we need to determine the values of
lim [x]
lim [x]
and
x→ –2 +
x→ –2 –
From (i), we see that
lim [x]
x→ –2 +
=
–2
Also, from the graph, we see that
lim [x]
=
–3
x→ –2 – The left and right hand limits are not equal, which means that
lim [x]
x→ –2
DOES NOT EXIST
(iii)
lim
x→–2.4
[x]
Like (ii), we need to evaluate the left and right hand limits:
lim
x→–2.4 +
[x]
lim [x]
and
x→–2.4 –
From the graph,
lim
x→–2.4 +
[x]
=
–3
=
–3
lim
and
x→–2.4 –
[x]
=
–3
which means
lim
x→–2.4
[x]
(B) If n is an integer, evaluate
lim
x→n–
[x]
lim
and
[x]
x→n+
One good approach to solving this problem is by determining the values of these limits, using selected values of n. In this case, we select n such that –10 ≤ x ≤ 10 . Thus we have the following table:
n
lim
[x]
n
lim
x→n–
x→n– –11
1
0
–9
–10
2
1
–8
–9
3
2
–7
–8
4
3
–6
–7
5
4
–5
–6
6
5
–4
–5
7
6
–3
–4
8
7
–2
–3
9
8
–1
–2
10
9
–10
n –10
lim
[x]
x→n+ –10
n
lim
x→n+ 1
1
–9
–9
2
2
–8
–8
3
3
–7
–7
4
4
–6
–6
5
5
–5
–5
6
6
–4
–4
7
7
–3
–3
8
8
–2
–2
9
9
–1
–1
10
10
[x]
From the first table, you can see that each value of
lim
x→n–
[x]
is ALWAYS 1 less than its corresponding value of n. Thus, we say that
lim
[x]
x→n–
=
n –1
Using this same technique, we evaluate
lim
x→n+
[x]
and we get the second table below.
[x]
The pattern is simply too clear: each value of
lim
x→n+
[x]
is equal to its corresponding value of n. This means
lim
x→n+
[x]
=
n
(C) From (B) above, we saw that, for any integer value of n (positive and negative),
lim
[x]
x→n–
=
n –1
(left hand limit)
and
lim
x→n+
[x]
=
n
(right hand limit)
Since the left and right hand limits are not equal, it follows that, for any integer n,
lim
x→n
[x]
DOES NOT EXIST
So now, we've seen that the limit does not exist for integer values. For what values of n does it exist then? Let's take one more look at the graph:
Observe the graph carefully, and you'll see that a limit does not exist for integer values. So, let's consider another category of numbers on this graph: non-integers. Take a number like 1.5. This number will have a left and right hand limit (which is 1). The same will apply to any non-integer. You can test this theory yourself. In the end, we find that the limit
lim
x→n
[x]
exists only for only non-integers. In other words, the limit exists if and only if n is a non-integer
Example 27 In the theory of relativity, the Lorentz contraction formula
L
=
Lo
√
1
v2 c2
–
expresses the length L of an object as a function of its velocity v with respect to an observer, where Lo is the length of the object at rest and c is the speed of light. Find
lim L
v→c– and interpret the result. Why is a left hand limit necessary?
Solution The function in question is
L
=
Lo
√
1
v2 c2
–
and we're given the following information: L = Object length V = Velocity of object Lo = Length of object at rest c = Speed of light (approx. 3.0 ×105km/s) The question is asking us to find the value of the limit
lim L
v→c–
Verbally speaking, we've been asked to determine what happens to the length of an object as its velocity approaches that of light. Mathematically, we want to know what happens to L as v→c from the left. Thus, from a logical perspective, if v→c– (which means v approaches c, but remains less than c), then
v2 c2
→
1
(i)
As v gets closer and closer to c, v2/c2 gets closer and closer to 1. Thus, we can assume that, at some point,
v2 c2
=
1
(ii)
which means
√
1
–
v2 c2
would approach zero
(iii)
Once again, it is assumed that we get to a point where
√
So,
lim L
v→c–
1
= = =
–
v2 c2
=
lim
Lo
v→c–
(iv)
0
√
1
lim
Lo
√1
lim
Lo
√0
v→c– v→c–
– – 1
v2 c2 From (i) and (ii)
= =
lim
Lo (0)
v→c–
=
0 lim– Lo v→c
From (iv)
0
Therefore,
lim L
= 0
v→c–
So, what is the significance of this result? First, don't forget that the function we're dealing with is a “contraction formula”. With that in mind, the value of the limit can be taken to mean that the length of the object approaches zero (i.e, the object “shrinks in length”) as its speed approaches the velocity of light. Here's another interpretation of the result, taken from Cramster: “As an object approaches the speed of light from velocities necessarily less than that of light, the object cotracts in the direction of motion to a limiting length of zero”.
Example 28 Is there a number a such that
3x2 + ax + a + 3
lim
x2 + x – 2
X→ –2
exists? If so, find the value of a and the value of the limit.
Solution Take a close look at the limit:
lim X→ –2
3x2 + ax + a + 3 x2 + x – 2
Clearly, we can't use the quotient rule or direct substitution, because then, it would mean the limit does not exist (as the denominator would evaluate to zero). For now, let's assume a limit exists. We'll represent the limit by L. Therefore,
lim X→ –2
3x2 + ax + a + 3 x2 + x – 2
=
L
(i)
Next, we try to simplify the limit. We start by factorizing the denominator:
lim X→ –2
3x2 + ax + a + 3 (x + 2) (x – 1)
=
L
(ii)
You'll find that, at this point, any method (quotient rule or direct substitution, although I prefer the latter) just might work. However, there's a problem: the expression (x + 2) in the denominator prohibits us from moving any further. We therefore partially eliminate the expression by taking it to the other side of the equation in (ii):
lim X→ –2
3x2 + ax + a + 3 (x – 1)
=
(x + 2)L
Now, we can evaluate the limit. Here we use direct substitution (i.e, we put x = –2)
(iii)
3(–2)2 + a(–2) + a + 3
=
(–2 – 1) 12 – 2a + a + 3
=
(–2 + 2)L
=
0
–3
15 – a
=
=
0
–3 Solving for a gives a = 15. Now that we know the value of a, we can find the value of L. To do that, we put a = 15 in (i) above:
3x2 + 15x + 15 + 3
lim
=
X→ –2
x2 + x – 2
lim
3x2 + 15x + 18 x2 + x – 2
X→ –2
=
L
=
L
We simplify the numerator and denominator to give
lim
3((x + 2)(x + 3))
X→ –2
(x + 2) (x – 1)
=
L
Canceling out common factors leaves
3(x + 3) (x – 1)
lim X→ –2
=
lim X→ –2
3x + 9 x–1
=
L
Finally, we use direct substitution to evaluate what's left of the limit (or you can use the limit laws. My advice:Use whatever method you're convenient with unless otherwise specified). In other words, we put x = –2:
3(–2) + 9 –2 – 1
=
–6 + 9 –2 – 1
=
–1
Thus, L = –1
Example 29 If
f(x)
=
x2 if x is rational 0
if x is irrational
prove that
lim f(x) = 0
x→0
Solution f is a function defined by two different functions in two different domains: ➔ f(x) = x2 if x is rational
(I)
➔ f(x) = 0 if x is irrational
(II)
Let's consider the first case:
f(x) = x2 for any rational value of x. Therefore,
lim f(x)
lim (x2)
=
x→0
x→0
=
02
=
(Law 9)
0
Next, let's look at the second case: f(x) = 0 for any irrational value of x. Thus,
lim f(x)
lim 0
=
x→0
(Law 7)
0
=
x→0
We see that, for both scenarios, the limits evaluate to zero. Thus, for ANY value of x, rational or irrational,
lim f(x) = 0
x→0
Example 30 Show by means of an example that
lim
x→a
[f(x) + g(x)]
may exist even though neither
lim f(x)
lim g(x)
nor
x→a
x→a
exists.
Solution To solve this problem, we apply one of the basic principles of problem solving – working backwards. In this regard, a feasible solution to this problem will be to come up with a function that can be broken down into two simpler functions whose limits do not exist at a specified number a. This is perhaps the toughest part. Here's a simple rational function I've come up with:
x2 + 5x + 6
h(x) =
x +3
This is the point where we start working backwards. Suppose we break h into two functions, m and n such that
h(x) = m(x) + n(x) where
x2 + 5x
m(x) =
x +3
and
n(x) =
6 x +3
Now, let's put a = – 3. We see that the limits
lim m(x)
x→ –3
lim n(x)
and
x→ –3
do NOT EXIST, because the denominators of both limits evaluate to zero; however, if we combine m and n, we get a different result:
lim m(x)
x→ –3
+
lim n(x)
x→ –3
=
=
lim
x→ –3
lim
x→ –3
[m(x) + n(x)] x2 + 5x x +3
+
6 x +3
x2 + 5x + 6
lim
=
x→ –3
x +3
=
lim
x→ –3
(x +2)(x +3 ) x +3
lim (x + 2)
=
x→ –3
=
–3 + 2
=
–1
(Direct Substitution)
Thus, even though the limits
lim m(x)
and
x→ –3
lim n(x)
x→ –3
did not exist, their sum yielded a value. The same applies to products of limits, as the next example will show. Try to practice this example over and over by deriving your own functions and proving the theorem.
Example 31 Show by means of an example that
[f(x)g(x)]
lim
x→a
may exist even though neither
lim f(x)
lim g(x)
nor
x→a
x→a
exists.
Solution We apply the “working backwards” technique we applied in Example 30 above. Here we use a relatively simple rational function:
h(x) =
2√x(3x2 – 15x) x√x(4 + 3x)
which is then split into two simpler functions:
f(x) =
2√x
and
x(4 + 3x)
g(x) =
3x2 – 15x √x
and put a = 0. Thus we have two limits
lim f(x)
x→0
and
lim g(x)
x→0
both of which do not exist because the denominators evaluate to zero. But if we combine the limits of f and g via multiplication, we get
lim
x→0
[f(x)g(x)]
=
=
=
lim f(x)
x→0
lim
× 2√x
x→0
x(4 + 3x)
lim
2√x
x→0
x(4 + 3x)
lim g(x)
x→0
×
3x2 – 15x
×
3x2 – 15x
√x
√x
=
lim
2√x(3x2 – 15x)
x→0
lim
=
x√x(4 + 3x)
x→0
2x√x(3x – 15) x√x(4 + 3x)
Canceling out common factors leaves
lim
2(3x – 15)
x→0
(4 + 3x)
=
6x – 30
lim
4 + 3x
x→0
You can see this limit can be evaluated without any problems. We use direct substitution:
6(0) – 30 4 + 3(0)
=
–30 4
=
–7.5
Again, we see that even though the limits
lim f(x)
x→0
lim g(x)
and
x→0
did not exist, their product yielded a value.
Example 32 Let f(x) = x – [x] (a) Sketch the graph of f. (b) If n is an integer, evaluate
lim f(x)
x→n–
lim f(x)
and
x→n+
(c) For what values does
lim f(x)
x→a exist?
Solution Observe that f contains a relatively familiar function, the greatest integer function y = [x]. We've examined this function in Example 26. To solve these problems, we need to apply what we've learnt about the function y = [x]. First, we have to graph f, and to do that, we need two vital pieces of information: ➔ How f is defined for integer values ➔ How f is defined for non-integer values Let's say we want to graph f for –5 ≤ x ≤ 5. In the first instance, we find that, for any integer x,
The solution to this example has been adapted from Cramster
f(x) = x – [x] = x–x
= 0
This means that f(x) = 0 if x is an integer. To determine how f is defined for non-integers, we select a few domains and see how f is defined in each domain, like this: ➔ In the domain –5 < x < –4 for example, [x] = –5 and therefore f(x) = x – [–5] = x – (–5) = x + 5 ➔ In the domain –4 < x < –3, [x] = –4 and therefore f(x) = x – [–4] = x – (–4) = x + 4 and so on. You'll understand the pattern better when its tabulated:
Domain of x
[x]
f(x) = x – [x]
–5 < x < –4
–5
f(x) = x + 5
–4 < x < –3
–4
f(x) = x + 4
–3 < x < –2
–3
f(x) = x + 3
–2 < x < –1
–2
f(x) = x + 2
–1 < x < 0
–1
f(x) = x + 1
0<x<1
0
f(x) = x
1<x<2
1
f(x) = x – 1
2<x<3
2
f(x) = x – 2
3<x<4
3
f(x) = x – 3
4<x<5
4
f(x) = x – 4
Using this table and the fact that f(x) = 0 for any integer x, we have the graph below.
(b)
From the graph below, it is quite obvious that, for instance, f(x)→1 as x→ –5 – , f(x)→1 as x→ 3 – , f(x)→1 as x→ 5 – , and so on, as illustrated by the upward- facing arrows in the graph. Thus, for any integer n,
lim f(x) = 1
x→n– On the other hand, we also see from the graph that, for example, f(x)→0 as x→ –5 +, f(x)→0 as x→ 3 +, f(x)→0 as x→ 5 +, and so on, as illustrated by the downward- facing arrows in the graph. Thus, for any integer n,
lim f(x) = 0
x→n+
y = [x]
(c) In summary, we've seen from (a) and (b) that, for any integer n,
lim f(x) = 1
x→n–
and
lim f(x) = 0
x→n+
Since the two one-sided limits are not equal, it follows that the limit
lim
f(x)
x→a does exists if a is an integer. However, if you look carefully at the graph, you'll see that a limit exists for non-integers, and this implies that the limit exists if and only if a is a non-integer.
Example 33 If f(x) = [x] + [– x], show that
lim
f(x)
x→2 exists but is not equal to f(2).
Solution If you read the question carefully, you'll see that, to solve the problem, we have to determine the values of ➔
lim
x→2
f(x)
➔ f(2) The question is asking us to establish the fact that
lim
x→2
f(x)
≠
f(2)
If f(x) = [x] + [– x], then
f(2) = [2] + [– 2] f(2) = 2 + (– 2) = 2 – 2 = 0 Thus, f(2) = 0. In fact, we see that, for ANY integer x, we see that f(x) = 0. To determine the value of
lim
f(x)
x→2 we apply the technique used to solve Example 32: evaluate f for different domains. Here, we choose the domain such that –2 < x < 3. Like Example 32, we use a table:
Domain of x
f(x) = [x] + [–x]
–2 < x < –1
f(x) = – 2 + 2 = – 1
–1 < x < 0
f(x) = – 1 + 0 = – 1
0<x<1
f(x) =
0 + (–1) = – 1
1<x<2
f(x) =
1 + (–2) = – 1
2<x<3
f(x) =
2 + (–3) = – 1
From the table, we see that, for any non-integer x, f(x) = –1. Using this information and the fact that f(x) = 0 for integer values of x, we have the following graph:
From the graph of f, it is quite clear that f(2) = 0, but
lim f(x) = –1
x→2 Thus, we have shown that
lim
x→2
f(x)
≠
f(2)
Example 34 Evaluate the limit, if it exists.
lim
x→1
√x – x2 1 – √x
Solution Observe that the numerator and denominator are irrational. Thus, this limit can be evaluated by rationalizing either. Here, we rationalize the denominator first:
lim
x→1
√x – x2 1 – √x
×
1 + √x
lim
=
1 + √x
(√x – x2)(1 + √x)
x→1
(1 – √x)(1 + √x)
lim
√x + x – x2 – x2√x
=
x→1
1–x
We need to cancel out the denominator, and doing this requires the rearrangement of the terms in the numerator:
lim
x→1
√x + x – x2 – x2√x 1–x
Then we factorize the numerator:
=
lim
x→1
√x – x2√x + x – x2 1–x
lim
√x – x2√x + x – x2
x→1
=
1–x
lim
x→1
lim
=
=
√x(1 – x2) + x(1 – x) 1–x √x(1 – x)(1 + x) + x(1 – x)
x→1
1–x
lim
√x(1 – x)(1 + x)
x→1
1–x
+
x(1 – x) 1–x
lim (√x(1 + x) + x )
=
x→1
lim (√x + x√x + x)
=
x→1
√(1) + (1)√(1) + (1)
=
3
= Therefore,
√x – x2
lim
=
1 – √x
x→1
3
Now, let's evaluate the limit by rationalizing the numerator:
lim
x→1
√x – x2 1 – √x
×
√x + x2
=
√x + x2
=
(√x – x2)(√x + x2)
lim
x→1
(1 – √x)(√x + x2)
lim
x – x4
x→1
(1 – √x)(√x + x2)
=
x(1 – x3)
lim
x→1
(1 – √x)(√x + x2)
After rationalizing the numerator, you can see that the limit hasn't been simplified much. In cases like this, we rationalize again using the original denominator:
lim
x→1
x(1 – x3) (1 – √x)(√x + x2)
×
1 + √x 1 + √x
=
=
=
=
lim
x→1
lim
x(1 – x3)(1 + √x) (1 – √x)(1 + √x)(√x + x2) –x(x3 – 1)(1 + √x)
x→1
(1 – x)(√x + x2)
lim
–x(x3 – 1)(1 + √x)
x→1
lim
–(x – 1)(√x + x2) x(x – 1)(x2 + x + 1)(1 + √x)
x→1
Crossing out common factors gives
lim
x→1
x(x2 + x + 1)(1 + √x) (√x + x2)
At this point, the limit has been simplified well enough for us to evaluate.
(x – 1)(√x + x2)
Using direct substitution,
lim
x(x2 + x + 1)(1 + √x)
x→1
(√x + x2)
=
= =
(1)((1)2 + (1) + 1)(1 + √(1)) (√(1) + (1)2) (1)(3)(2) (2)
3
You'll agree that the second method is relatively complicated, but I want you to realize that sometimes, you need to rationalize a relatively complicated rational limit more than once before you can move any further.
The solutions to example 34 have been adapted from Cramster
Example 35 The figure shows a fixed circle C1 with equation (x – 1)2 + y2 = 1 and a shrinking circle C2 with radius r and center the origin. P is the point (0, r), Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r → 0+ ? y
C2
P
Q
0
R
x
C1 Note: Figure not drawn to scale
Solution The figure shows two circles, C1, whose equation is (x – 1)2 + y2 = 1. This equation can be rewritten to look like a standard circle equation:
(x – 1)2 + y2 = 1 =
x2 – 2x + 1 + y2 = 1
=
x2 – 2x + y2 = 1 – 1
=
x2 – 2x + y2 = 0
=
x2 + y2 = 2x
(i)
There's a second circle C2 whose equation is not given. Nonetheless, we know that it has a radius r and its center is the origin (0,0). Thus we use the standard circle equation:
x2 + y2 = r2
(ii)
to represent C2. The task here is to determine what happens to R as r → 0+ . Try to imagine for a moment what the figure would look like when C2 starts to shrink! You'll find that the line PQ starts to shrink too, and this means that the point R begins to move as well. To be more specific, the x-coordinate of R will approach a certain value. But what is the limit? Well, we're about to find out. We start by solving (i) and (ii) simultaneously:
x2 + y2 = 2x
(i)
x2 + y2 = r2
(ii)
When we solve the two equations, we find that r2 = 2x. Solving for x gives
x
r2 2
=
(iii)
This will be the x-coordinate of the point Q (where C 1 and C2 intersect). To get the y-coordinate, we plug (iii) into (ii) to give 2
r2 2 r4 4
=
+
+
y2
y2
=
r2
=
r2
Therefore,
y2
=
y2
=
y
=
√
y
=
r 2
r4 4
r2 – 4r2 – r4 4
r2(4 – r2) 4
=
So,
r2(4 – r2) 4
√
=
r2 (4 – r2) 4
√(4 – r ) 2
This is the y-coordinate of Q. Therefore, Coordinate of point Q
r2 , 2
=
r 2
√(4 – r ) 2
and from the graph, the coordinate of P = (0, r). Now that we have the coordinates of two points P and Q, we can compute the slope m:
Slope of PQ
=
=
=
r 2
√(4 – r )
–
2
r2 2
–
r
=
√(4 – r )
–
1
2
1 2
√(4 – r )
–
1
r
–
2
r
r2 2
1 2
2
√(4 – r )
0
r
2
r 2
×
2 r2
√(4 – r ) – 2
=
r
2
Therefore,
√(4 – r ) – 2
m
=
2
r
We have the slope of PQ, which means we can compute the equation of the line. Using the slope-intercept equation of a line, we have
y
=
mx + c
where m = slope of PQ c = y-intercept (which in this case, equals r) Thus the equation can be rewritten as
y
=
mx + r
(iv)
Recall the problem, we want to know what happens to the point R (x, 0) as r → 0+ . We see that if r → 0+ , then the point R begins to shift. To be more specific, the x-coordinate of R changes. But what point does it approach? That's what we really want to know. Therefore, we'll need an equation that explicitly describes the x-coordinate of R. In other words, we express (iv) in terms of r, which is done by making x the subject of (iv): We know that R = (x,y) = (x,0). Putting y = 0 in (iv) gives
mx + c ∴
= 0
–r m
x =
(v)
Plugging in the value of m into (iv) gives
–r x
=
√(4 – r ) – 2
2
r
Which equals
x
=
– r2
√(4 – r ) – 2
(vi)
2
We are getting close!! Once again, recall the problem – we want to know what happens to the point R (x, 0) as r → 0+ . The answer can be found by evaluating the limit
– r2
lim
r→0 +
√(4 – r ) – 2
2
Since the denominator is irrational, we rationalize it. This yields
lim
– r2
r→0 +
=
√(4 – r ) – 2
– r2
lim
r→0
2
×
√(4 – r ) √(4 – r )
+ +
2 2
√(4 – r ) 2
+
2 2
2
+
√(4 – r ) – 2
2
√(4 – r ) 2
+
2
=
=
=
lim
r→0
+
lim
r→0
√(4 – r )
– r2
lim
2
4 – r2 – 4
– r2
√(4 – r )
+
2
+
r→0 +
+
2
2
– r2
√(4 – r )
+
2
2
Using direct substitution,
lim
r→0 +
√(4 – r )
+
2
2
=
√
=
√4 + 2
=
2+2 4
=
+
4 – (0)2
2
Thus,
– r2
lim
√(4 – r ) –
r→0 +
2
= 2
4
In other words, as r → 0+ , the x-coordinate of R approaches 4. Here's a better interpretation: as r → 0+, the point R approaches the point (4,0) on the x-axis.
The solution to example 35 has been adapted from Cramster
Example 36 Evaluate the limit
lim
x→2
√6 √3
– x
–
2
– x
–
1
Solution I've tried to evaluate this limit using many methods. I came to understand that the best (and easiest) way around it is to “double rationalize”! See how it's done:
lim
x→2
√6 √6
√3 √3
√6 √3
– x
–
2
– x
–
1
√6
– x
–
2
√6
– x
+
2
√3
– x
+
1
√3
– x
–
1
√6
– x
+
2
√3
– x
+
1
×
– x
+
2
– x
+
2
×
which equals
lim
x→2
– x
+
1
– x
+
1
lim
=
x→2
(6
– x – 4)
√3
– x
+
1
(3
– x – 1)
√6
– x
+
2
(2
– x)
√3
– x
+
1
(2
– x)
√6
– x
+
2
lim
=
x→2
Crossing out common factors gives
√3 – x + √6 – x +
lim
x→2
1 2
The function is still irrational, but at this point, it does not matter because it can be evaluated That's the whole point of rationalization – to simplify a function far enough so that it can be evaluated as easily as possible. So now, we use direct substitution:
lim
x→2
√3 – x + √6 – x +
1
√3 – 2 + √6 – 2 +
=
2 =
√1 + 1 √4 + 2
=
1 2
1 2
Therefore,
√6 √3
lim
x→2
– x
–
2
– x
–
1
=
1 2
The solution to example 36 has been adapted from Cramster
This tutorial is concluded in Part II, where we look at a very important Theorem: The Sandwich Theorem. Before you exit, try the following exercises. If you need help, you know how to reach me.
EXERCISES (1)
Given that
lim f(x) = –3
x→a
lim g(x) = 0
x→a
lim h(x) = 8
x→a
find the limits that exist. If the limit does not exist, explain why.
(a) (b) (c)
lim [f(x) + h(x)]
x→a
lim
x→a
[f(x)]2
lim [h(x)]1/3
x→a
lim
(d)
x→a
(e)
x→a
(f)
1 f(x)
f(x) h(x) lim g(x) x→a f(x)
lim
(g)
f(x) x→a g(x)
(h)
2f(x) h(x) – f(x) x→a
lim
lim
(2)
Evaluate the limits and justify each step by indicating the appropriate limit laws.
x – 2 x→–1 x + 4x – 3
lim
(a)
x→4 –
4 lim (1 + h) – 1
(d)
x2 – 81 √x – 3
lim
(g)
x→9
(h)
lim
x→0
x2 – 2x
x→0
lim
(l)
x→9 +
(f)
x√1 + x
–
(n)
x2 + x – 6 x2 – 4
lim
x→2
1
(i)
x
lim |x + 4| lim
x→2
)
(√x – 9
x2 – x – 12 x+3
x→– 4
√x + 2 – √2x
lim
(k)
1
(m)
h
h→0
x→–3
h
h→0
2 lim 1 – √1 – h
(j)
lim
(c)
√16 – x2
3 lim (2 – h ) – 8
(e)
h
h→0
lim
(b)
2
|x – 2| x –2
+ [x + 1]
(3)(a) Estimate the value of x
lim
√1 + 3x –
x→0
1
by graphing the function
f(x)
x
=
√1 + 3x –
1
(b) Make a table of values for f(x) for x close to 0 and guess the value of the limit. (c) Use the limit laws to prove that your guess is correct . (4)(a) Use a graph of f(x)
=
√3 + x – √3 x
f(x) to two decimal places
to estimate the value of limx→0
(b) Use a table of values of f(x) to estimate the limit to four decimal places (c) Use the limit laws to find the exact value of the limit. (5)
Let
if x < 0
√–x f(x) =
3–x
if 0 ≤ x ≤ 3 if x > 3
(x – 3)
2
Evaluate each limit, if it exists.
(a)
lim f(x)
x→0
(b)
lim f(x)
x→3
calculus4engineeringstudents.com
lim
x→16
4 – √x x – 16