ThE LIMIt OF A FUNCTION THE SANDWICH THEOREM In the previous section, we examined eleven limit laws and have illustrated each law with several examples. Note that the limit laws are also the properties of limits. Before we conclude this tutorial, we take a look at two additional properties of limits (or limit laws), expressed as Theorems.
If f(x) ≤ g(x) when x is close to, but NOT equal to a and the limits of f and g both exist, then,
lim f(x)
x→a
≤
lim g(x)
x→a
Theorem 1 The next law is based on Theorem 1, and is called the Sandwich Theorem. This Theorem is stated thus:
If f(x) ≤ g(x) ≤ h(x) when x is close to, but NOT equal to a and
then
lim f(x)
=
lim g(x)
=
x→a
x→a
lim h(x)
x→a
=
L
L Theorem 2
You're probably wondering why this Theorem is called a “Sandwich” Theorem. Perhaps this graph would partly explain why:
On the graph, there are three different functions, f, g and h. The function whose limit we really want to find is g. Observe that g appears to be “sandwiched between f and h near the origin. Since the limits of f and h exist at a and are equal, then g is “forced” to have the same limit. In the graph above for instance, f and h have the same limit (which is zero) at a = 0; and since g is “sandwiched” between f and h, it's forced to have a limit of zero. Alternatively, the Theorem can be called the Squeeze Theorem or Pinching Theorem. Although my textbook uses the word “Squeeze”, I prefer to use “Sandwich” because it's more illustrative of the Theorem is really about. Let's illustrate the Theorem with a few examples and at the end, you get to try a couple of exercises.
Example 1 If 3x ≤ f(x) ≤ x3 + 2 for 0 ≤ x ≤ 2, evaluate
lim f(x)
x→1
Solution Observe that the question presents three different functions: ➔ m(x) = 3x ➔ f(x) ➔ n(x) = x3 + 2 Note that f is not explicitly defined, which means we'll concentrate on the other two for now. If we are to go by the Sandwich Theorem,
lim m(x)
≤
lim (3x)
≤
x→1
=
x→1
lim f(x)
≤
lim f(x)
≤
x→1
x→1
lim n(x)
x→1
lim (x3 + 2)
x→1
Next, we evaluate the two limits via direct substitution:
3(1)
≤
3
≤
=
lim f(x)
≤
lim f(x)
≤
x→1
x→1
(1)3 + 2 3
The two limits
lim (3x)
x→1
lim (x3 + 2)
and
x→1
evaluated to 3 and since f is “sandwiched between m and n, then by the Sandwich Theorem,
lim f(x) =
x→1
3
A graphical illustration is shown on the next page.
Example 2 If 1 ≤ f(x) ≤ x2 + 2x + 2 for all x, evaluate
lim f(x)
x→ –1
Solution Again, we have three functions, f, g and h, defined for all x. From the Sandwich Theorem,
lim
x→-1
1
lim f(x)
≤
x→-1
lim
≤
x→-1
(x2 + 2x + 2)
Using direct substitution and Law 7 (the limit of a constant),
=
1
≤
lim f(x)
≤
((–1) 2 + 2(–1) + 2)
1
≤
lim f(x)
≤
1
x→-1
x→-1
Observe that the limits on either side of
lim f(x)
x→ –1
evaluated to 1. Since f is squeezed between the other two functions, which have the same limit at a = –1 (which is 1), then, by the Sandwich Theorem,
lim f(x) = 1
x→ –1
Example 3 Use the Squeeze Theorem to show that
lim
x→0
√x
3
+ x2 sin
π x
=
0
Illustrate by graphing the functions f, g and h (in the notation of the Squeeze Theorem) on the same screen.
Solution Observe the function carefully, and you'll see that
√x
3
+ x2 sin
π x
=
+ x2 sin
π x
=
√x
3
×
sin
+ x2
×
+ x2
π x
which means
lim
x→0
√x
3
lim
x→0
√x
3
lim sin π
x→0
x
There's a small problem here: the limit
lim
x→0
sin
π x
does not exist. However, there's another vital clue that must be put into consideration:
–1 as shown from its graph:
≤
sin
π x
≤
1
(ii)
(i)
Now that we've established (ii), we multiply both sides of the double inequality by
√x
+ x2
√x
+ x2
3
to give
–1
=
lim
x→0
–
√x
3
3
≤
lim
≤
+ x2
x→0
√x
3
√x
3
+ x2
+ x2
sin
sin
π x
≤
π x
≤
1
√x
lim
x→0
3
+ x2
√x
3
+ x2
(iii)
Comparing (iii) with the Squeeze Theorem notation, we see that
√x
f(x)
=
–
g(x)
=
√x
h(x)
=
3
+ x2
+ x2
3
√x
3
sin
π x
+ x2
Graphing f, g, and h gives
From (iii) and the graph above, we see that f and h have the same limit at 0 (which is also zero). The graph clearly shows that g is sandwiched between f and h, which means that g is also forced to have a limit of zero. In other words,
lim
x→0
√x
3
+ x2 sin
π x
=
0
Example 4 Prove that
lim
x→0
x4 cos
2 x
=
0
Solution We find that
lim
x→0
x4 cos
2 x
=
lim x4
×
x→0
lim cos 2
(i)
x
x→0
Observe that
lim cos 2
x
x→0
DOES NOT EXIST. However, we also find that
–1
≤
cos
2 x
(ii)
1
≤
as shown by its graph:
Multiplying both sides of (ii) by x4 gives
=
– 1(x4)
≤
(x4) cos
2 x
≤
1(x4)
– x4
≤
x4 cos
2 x
≤
x4
=
lim
– x4
x→0
lim
≤
x→0
x4 cos
2 x
≤
lim
x→0
Using the Squeeze Theorem notation, we have three functions from (iii) above:
f(x)
=
– x4
g(x)
=
x4 cos
h(x)
=
2 x
x4
Again, we graph f, g, and h. Observe carefully:
From the graph above and (iii), we find that
lim f(x)
x→0
=
lim h(x)
x→0
and since g is sandwiched between f and h, then
lim
x→0 by the Squeeze Theorem. Let's try one more example.
x4 cos
2 x
=
0
=
0
x4
(iii)
Example 5 Prove that
lim
x→0 +
√ x [1
]
+ sin2(2π/x)
=
0
Solution Like examples 3 and 4, we see that we're dealing with a product:
lim
x→0 +
√ x [1
]
lim
=
+ sin2(2π/x)
x→0 +
√x
×
lim
x→0 +
[1
]
+ sin2(2π/x)
(i)
Observe that the limit
lim
x→0 +
[1
]
+ sin2(2π/x)
DOES NOT EXIST, but we see from the graph of y = 1 + sin2(2π/x) that
1
≤
[1
]
≤
+ sin2(2π/x)
(ii)
2
Multiplying (ii) by √x gives
[
]
≤
2(√x )
[
]
≤
2√x
]
≤
1(√x )
≤
(√x ) 1 + sin2(2π/x)
=
√x
≤
(√x ) 1 + sin2(2π/x)
=
x→0 +
lim √x
≤
lim
[
2 x→0 + (√x ) 1 + sin (2π/x)
lim 2√x
x→0 +
(iii)
So, by the notation of the the Squeeze Theorem,
f(x)
=
√x
g(x)
=
(√x ) 1 + sin2(2π/x)
h(x)
=
2√x
[
]
Graphing f, g, and h gives
From (iii) and the graph above,
lim f(x)
lim h(x)
=
x→0 +
=
x→0 +
0
Thus (iii) becomes
0
[
lim
≤
]
2 x→0 + (√x ) 1 + sin (2π/x)
≤
Since g is squeezed between f and h, then from (iii) and (iv),
lim
x→0 +
√ x [1
which illustrates the Sandwich Theorem. Try the following exercises:
]
+ sin2(2π/x)
=
0
0
(iv)
EXERCISE 1 Use the Squeeze Theorem to show that
lim x2 cos 20πx
x→0
=
Illustrate by graphing the functions f(x) = –x2,
0 g(x) = x2 cos 20πx, and h(x) = x2 on the same screen.
EXERCISE 2 If 2x – 1 ≤ f(x) ≤ x2 for 0 < x < 3, evaluate
lim f(x)
x→1
EXERCISE 3 Prove that
lim x2 cos (1/x2)
x→0
=
0
In the next tutorial, we examine CONTINUITY. In the previous tutorial, we saw that the limits many functions could be evaluated via direct substitution. Such functions exhibit CONTINUITY and are said to be CONTINUOUS. See how this property really works in the next tutorial.
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