# Indeterminate forms of Limits

Examples with detailed solutions and exercises that solves limits questions related to indeterminate forms such as :

## Theorem

A second version of L'Hopital's rule allows us to replace the limit problem

with another simpler problem to solve.

### Example 1

Find the limit $$\lim_{x\to \infty} \dfrac{\ln x}{x}$$
Solution to Example 1:
Since
$$\lim_{x\to \infty} \ln x = \infty$$
and
$$\lim_{x\to \infty} x = \infty$$
we have the indeterminate form
$$\lim_{x\to \infty} \dfrac{\ln x}{x} = \dfrac{\infty}{\infty}$$
The above L'Hopital's rule can be used to evaluate the given limit question as follows
$$\lim_{x\to \infty} \dfrac{\ln x}{x} = \lim_{x\to \infty} \dfrac{ (\ln x)' }{(x)'}$$
Evaluate the derivatives in the numerator and the denominator
$$= \lim_{x\to \infty} \dfrac{ 1/x }{1}$$
Evaluate the limits in the numerator and denominator
$$\lim_{x\to \infty} (1/x) = 0$$ and $$\lim_{x\to \infty} 1 = 1$$
We now evaluate the given limit
$$\lim_{x\to \infty} \dfrac{\ln x}{x} = \lim_{x\to \infty} \dfrac{ 0 }{1} = 0$$

### Example 2

Find $$\lim_{x\to \infty} x e^{-x}$$
Solution to Example 2:
Note that
$$\lim_{x\to \infty} x = \infty$$
and
$$\lim_{x\to \infty} e^{-x} = 0$$
This is the indeterminate form $$\infty \cdot 0$$. The idea is to convert it into to the indeterminate form $$\dfrac{\infty}{\infty}$$ and use L'Hopital's theorem. Note that
$$\lim_{x\to \infty} x e^{-x} = \lim_{x\to \infty} \dfrac{ x}{ e^x} = \dfrac{ \infty }{ \infty }$$
We apply the above L'Hopital's theorem
$$= \lim_{x\to \infty} \dfrac{ (x)'}{ (e^x)'} = \lim_{x\to \infty} \dfrac{ 1 }{ e^x} = 0$$

### Example 3

Find $$\lim_{x\to \infty} ( 1 + 1/x)^x$$

Solution to Example 3:
Note that $$\lim_{x\to \infty} (1 + 1/x) = 1$$ and the above limit is given by
$$\lim_{x\to \infty} ( 1 + 1/x)^x = 1^{\infty}$$
which is of the Indeterminate form $$1^{\infty}$$. If we let $$t = 1 / x$$ the above limit may written as
$$\lim_{x\to \infty} ( 1 + 1/x)^x = \lim_{t\to 0} ( 1 + t)^{1/t}$$ , note that as $$x\to \infty$$ , $$t\to 0$$
Let $$y = ( 1 + t)^{1/t}$$ and find the limit of $$\ln y$$ as t approaches 0
$$\ln y = \ln ( 1 + t)^{1/t} =(1 / t) \ln (1 + t)$$
The advantage of using $$\ln y$$ is that
$$\lim_{t\to 0} \ln y = \lim_{t\to 0} \dfrac {\ln (1 + t)}{t} = \dfrac{0}{0}$$
and the limit has the indeterminate form 0 / 0 and the first L'Hopital's rule can be applied
$$\lim_{t\to 0} \ln y = \lim_{t\to 0} \dfrac {ln (1 + t)}{t}$$
$$= \lim_{t\to 0} \dfrac{(ln (1 + t))'}{(t)'} = \lim_{t\to 0} \dfrac{1/(1+t)}{1} = 1$$
Since the limit of ln y = 1 the limit of y is e 1 = e, hence
$$\lim_{x\to \infty} ( 1 + 1/x)^x = e$$

### Example 4

Find the limit $$\lim_{x\to 0^+} ( 1 / x - 1 / \sin x )$$
Solution to Example 4:
Note that
$$\lim_{x\to 0^+} ( 1 / x ) = + \infty$$
and
$$\lim_{x\to 0^+} ( 1 / \sin x ) = + \infty$$
This limit has the indeterminate form $$\infty - \infty$$ and has to be converted to another form by combining $$1 / x - 1 / sin x$$
$$\lim_{x\to 0^+} ( 1 / x - 1 / \sin x ) = \lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \dfrac{0}{0}$$
We now have the indeterminate form 0 / 0 and we can use the L'Hopital's theorem.
$$\lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \lim_{x\to 0^+} \dfrac{(\sin x - x)'}{(x \sin x)'} = \lim_{x\to 0^+} \dfrac{\cos x - 1}{ \sin x + x \cos x} = \dfrac{0}{0}$$
We have again the indeterminate form 0 / 0 and use the L'Hopital's theorem one more time.
$$= \lim_{x\to 0^+} \dfrac{(\cos x - 1)'}{ (\sin x + x \cos x)'}$$
$$= \lim_{x\to 0^+} \dfrac{-\sin x}{ \cos x + \cos x - x \sin x } = \dfrac{0}{2} = 0$$

### Example 5

Find the limit $$\lim_{x\to 0^+} x^x$$
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
$$\lim_{x\to 0^+} \ln y = \lim_{x\to 0^+} x \ln x = 0 \cdot \infty$$
The above limit has the indeterminate form $$0 \cdot \infty$$. We have convert it as follows
$$\lim_{x\to 0^+} x \ln x = \lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \dfrac{\infty}{\infty}$$
It now has the indeterminate form $$\dfrac{\infty}{\infty}$$ and we can use the L'Hopital's theorem
$$\lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \lim_{x\to 0^+} \dfrac{(\ln x)'}{(1/x)'}$$
$$= \lim_{x\to 0^+} \dfrac{1/x}{-1/x^2}$$
$$= \lim_{x\to 0^+} - x = 0$$
The limit of ln y = 0 and the limit of y = xx is equal to
$$e^0 = 1$$

## Exercises

Find the limits
1. $$\lim_{x\to \infty} (\ln x)^{1/x}$$
2. $$\lim_{x\to \infty} (\ln x - \ln (1 + x))$$
3. $$\lim_{x\to \infty} \dfrac{x}{e^x}$$
4. $$\lim_{x\to 0^+} x^{\sin x}$$

1. 1
2. 0
3. 0
4. 1