Indeterminate forms of Limits

Examples with detailed solutions and exercises that solves limits questions related to indeterminate 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 evaluate the given limit question
limx→∞ ln x / x = limx→∞ [ d ( ln x ) / dx ] / [ d ( x ) / dx ]
= limx→∞ [ 1 / x ] / 1 = 0

Example 2: Find 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 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 form 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

More on limits
Calculus Tutorials and Problems

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