Examples with detailed solutions and exrcises that solves limits questions related to inderminate forms such as
¥ / ¥, 0 0, ¥ 0, 1 ¥, ¥ o and ¥ - ¥.
A second version of L'hopital's rule allows us to replace the limit problem ¥ / ¥ with another simpler problem to solve.
Theorem: If lim f(x) = ¥ and lim g(x) = ¥ and if lim [ f'(x) / g'(x) ] has a finite value L , or is of the form + ¥ or - ¥, then
lim [ f(x) / g(x) ] = lim [ f'(x) / g'(x) ]
lim stands for limx® a, limx® a+, limx® a-, limx® + ¥ or limx® - ¥.
Example 1: Find the limit limx® +¥ ln x / x
Solution to Example 1:
Since
limx® +¥ ln x = +¥
and
limx® +¥ x = +¥
The above L'hopital's rule can be used to evalute the given limit question
limx® +¥ ln x / x = limx® +¥ [ d ( ln x ) / dx ] / [ d ( x ) / dx ]
= limx® +¥ [ 1 / x ] / 1 = 0
Example 2: Find the limit limx® + ¥ x e -x
Solution to Example 2:
Note that
limx® + ¥ x = + ¥
and
limx® + ¥ e - x = 0
This is the indeterminate form ¥ . 0. The idea is to convert it into to the indeterminate form ¥ / ¥
limx® + ¥ x e -x
= limx® + ¥ x / e x
Now apply the above L'hopital's theorem
limx® + ¥ x / e x = limx® + ¥ 1 / e x = 0
Example 3: Find the limit limx®+ ¥ ( 1 + 1/x) x
Solution to Example 3:
The above limit is of the indeterminate form 1 ¥. If we let t = 1 / x the above limit may written as
limx®+ ¥ ( 1 + 1/x) x = limt® 0 ( 1 + t) 1 / t
Let y = ( 1 + t) 1 / t and find the limit of ln y as t approaches 0
ln y = (1 / t) ln (1 + t)
The advantage of using ln y is that now the limit has the indeterminate for 0 / 0 and the first L'hopital's rule can be applied
limt® 0 (1 / t) ln (1 + t) = limt® 0 [ d ( ln(1 + t) ) / dt ] / [ d ( t ) / dt ]
= limt® 0 [ 1 / (1 + t) / 1 = 1
Since the limit of ln y = 1 the limit of y is e 1 = e, hence
limx®+ ¥ ( 1 + 1 / x) x = e
Example 4: Find the limit limx® 0+ (1 / x - 1 / sin x)
Solution to Example 4:
Note that
limx® 0+ 1 / x = + ¥
and
limx® 0+ 1 / sin x = + ¥
This limit has the indeterminate form ¥ - ¥ and has to be converted to another form by combining 1 / x - 1 / sin x
limx® 0+ (1 / x - 1 / sin x) = limx® 0+ [ (sin x - x) / (x sin x) ]
We now have the indeterminate form 0 / 0 and we can use the L'hopital's theorem.
limx® 0+ [ (sin x - x) / (x sin x) ]
= limx® 0+ [ (cos x - 1) / (sin x + x cos x) ]
We have the indeterminate form 0 / 0 and use the L'hopital's theorem again.
limx® 0+ [ (sin x - x) / (x sin x) ]
= limx® 0+ [ (cos x - 1) / (sin x + x cos x) ]
= limx® 0+ [ (-sin x) / (cos x + cos x - x sin x) ] = 0 / 2 = 0
Example 5: Find the limit limx® 0+ x x
Solution to Example 5:
We have the indeterminate form 00. Let y = x x and ln y = ln (x x) = x ln x. Let us now find the limit of ln y
limx® 0+ ln y
= limx® 0+ x ln x
The above limit has the indeterminate form 0 . ¥. We have convert it as follows
limx® 0+ x ln x
= limx® 0+ ln x / (1 / x)
It now has the indeterminate form ¥ / ¥ and we can use the L'hopital's theorem
limx® 0+ ln x / (1 / x)
= limx® 0+ (1 / x) / (- 1 / x 2)
= limx® 0+ -x = 0
The limit of ln y = 0 and the limit of y = xx is equal to
e 0 = 1
Exercises: Find the limits
1. limx® +¥ (ln x) 1/x
2. limx® +¥ [ ln x - ln (1 + x) ]
3. limx® +¥ x / e x
4. limx® 0+
x sin x
Solutions to Above Exercises: Find the limits
1. 1
2. 0
3. 0
4. 1
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