Solved Problems on Limits of Functions
Problems 1
x 2 − 3x + 2 lim x →2 x −2
2
x3 + x2 + x + 1 lim 3 x →∞ x + 3 x 2 + 5 x + 2
3
lim x 2 + 1 − x 2 − 1
4
lim x 2 + x + 1 − x 2 − x − 1
5
x →∞
x →∞
lim x →0
2x 2 x 2 + x + 1 − x 2 − 3x + 1 Mika Seppälä: Solved Problems on Limits of Functions
Main Methods of Limit Computations 1
2
The following undefined quantities cause problems: 0 ∞ 00 , , , ∞ − ∞, 0∞ , ∞0. 0 ∞ In the evaluation of expressions, use the rules
a ∞ = 0, = ∞, ∞ × (negative number ) = −∞. ∞ positive number 3
4
If the function, for which the limit needs to be computed, is defined by an algebraic expression, which takes a finite value at the limit point, then this finite value is the limit value. If the function, for which the limit needs to be computed, cannot be evaluated at the limit point (i.e. the value is an undefined expression like in (1)), then find a rewriting of the function to a form which can be evaluated at the limit point. Mika Seppälä: Solved Problems on Limits of Functions
Main Computation Methods 1 2
3
Frequently needed rule
( a + b ) ( a − b ) = a2 − b2.
Cancel out common factors of rational functions. x 2 − 1 ( x − 1) ( x + 1) = = x + 1 ⎯⎯⎯⎯ → 2. x →1 x −1 x −1 If a square root appears in the expression, then multiply and divide by the conjugate of the square root expression.
x +1 −
( x −2 =
x +1 −
x +2
)(
x +1 + ( x + 1) − ( x − 2 ) = = x +1 + x −2
x +1 +
x −2
)
x −2 3 x +1 +
x +2
Mika Seppälä: Solved Problems on Limits of Functions
⎯⎯⎯⎯ →0 x →∞
Limits by Rewriting Problem 1
Solution
x 2 − 3x + 2 lim x →2 x −2 x 2 − 3 x + 2 ( x − 1)( x − 2 ) Rewrite = = x − 1. x −2 x −2 x 2 − 3x + 2 Hence lim = lim ( x − 1) = 1. x →2 x →2 x −2
Mika Seppälä: Solved Problems on Limits of Functions
Limits by Rewriting Problem 2
x3 + x2 + x + 1 lim 3 x →∞ x + 3 x 2 + 5 x + 2
Solution
1 + x3 + x2 + x + 1 x = x 3 + 3x 2 + 5x + 2 1+ 3 + x 1+
1 1 + x 2 x 3 ⎯⎯⎯→ 1. x →∞ 5 2 + 3 2 x x
Mika Seppälä: Solved Problems on Limits of Functions
Limits by Rewriting lim x 2 + 1 − x 2 − 1
Problem 3 Solution
x →∞
Rewrite
x +1− x 2
( =
2
( −1 =
) ( 2
x +1 − 2
)(
x2 + 1 − x2 − 1
x2 + 1 + x2 − 1
)
x −1 2
x2 + 1 + x2 − 1
2
x ( =
2
) (
x →∞
)=
+ 1 − x2 − 1
x2 + 1 + x2 − 1
Hence lim x + 1 − x − 1 = lim 2
)
x2 + 1 + x2 − 1
x2 + 1 + x2 − 1
2
2
x →∞
2
x +1+ x −1 2
2
Mika Seppälä: Solved Problems on Limits of Functions
= 0.
Limits by Rewriting lim x 2 + x + 1 − x 2 − x − 1
Problem 4
x →∞
Rewrite
Solution
x2 + x + 1 − x2 − x − 1 =
( x ( =
2
)
x + x +1− x − x −1 2
) (
2
)=
+ x + 1 − x2 − x − 1
x2 + x + 1 + x2 − x − 1 x2 + x + 1 + x2 − x − 1 2x + 2
x2 + x + 1 + x2 − x − 1 x2 + x + 1 + x2 − x − 1 2 2+ Next divide by x. x = ⎯⎯⎯ →2 x →∞ 1 1 1 1 1+ + 2 + 1− − 2 x x x x Mika Seppälä: Solved Problems on Limits of Functions
Limits by Rewriting Problem 5
lim
x →0
2x 2x 2 + x + 1 − x 2 − 3 x + 1
Rewrite
Solution
2x
=
2x + x + 1 − x − 3 x + 1 2
2
2x
(
(
)
2x 2 + x + 1 + x 2 − 3 x + 1
)(
2x 2 + x + 1 − x 2 − 3 x + 1
2x ( ) = = ( 2x + x + 1) − ( x − 3x + 1) 2 ( 2 x + x + 1 + x − 3 x + 1) = ⎯⎯⎯ →1 2x
(
2x 2 + x + 1 + x 2 − 3 x + 1 2
2
2
2
2
2
x+4
)
2x 2 + x + 1 + x 2 − 3 x + 1
)
2x 2 + x + 1 + x 2 − 3 x + 1 x 2 + 4x Next divide by x.
x →0
Mika Seppälä: Solved Problems on Limits of Functions