# L'Hopital's Rule and The Indeterminate Forms of Limits in Calculus

L'Hopital's theorem allows us to replace a limit problem with another that may be simpler to solve. Several examples are presented along with their solutions and detailed explanations.

## L'Hopital's Theorem

If $$\lim_{{x \to a}} f(x) = 0$$ and $$\lim_{{x \to a}} g(x) = 0$$ and if $$\lim_{{x \to a}} \dfrac{{f'(x)}}{{g'(x)}}$$ has a finite value $$L$$, or is of the form $$+\infty$$ or $$-\infty$$, then

where $$\lim$$ stands for $$\lim_{{x \to a}}, \lim_{{x \to a^+}}, \lim_{{x \to a^-}}, \lim_{{x \to +\infty}},$$ or $$\lim_{{x \to -\infty}}$$.
$$f'(x)$$ and $$g'(x)$$ are the derivatives of $$f(x)$$ and $$g(x)$$ respectively.

## Examples with Detailed Solutions

### Example 1

Find the limit $$\lim_{{x \to 0}} \dfrac{{\sin x}}{{x}}$$

Solution to Example 1:
Since
$$\lim_{{x \to 0}} \sin x = 0$$
and
$$\lim_{{x \to 0}} x = 0$$
L'Hopital's rule can be used to evaluate the above limit as follows
$$\lim_{{x \to 0}} \dfrac{{\sin x}}{{x}} = \lim_{{x \to 0}} \dfrac{\dfrac{{d(\sin x)}}{{dx}}} {\dfrac{{d(x)}}{{dx}} }$$
$$= \lim_{{x \to 0}} \dfrac{{\cos x}}{{1}} = \dfrac{{\cos 0}}{{1}} = 1$$

### Example 2

Find the limit $$\lim_{{x \to 0}} \dfrac{{e^x - 1}}{{x}}$$

Solution to Example 2:
Note that
$$\lim_{{x \to 0}} (e^x - 1) = 0$$
and
$$\lim_{{x \to 0}} x = 0$$
We can use L'Hopital's rule to calculate the given limit as follows
$$\lim_{{x \to 0}} \dfrac{{e^x - 1}}{{x}} = \lim_{{x \to 0}} \dfrac{\dfrac{{d(e^x - 1)}}{{dx}}} { \dfrac{{d(x)}}{{dx}}}$$
$$= \lim_{{x \to 0}} \dfrac{{e^x}}{{1}} = \dfrac{{e^0}}{{1}} = 1$$

### Example 3

Find the limit $$\lim_{{x \to 1}} \dfrac{{x^2 - 1}}{{x - 1}}$$

Solution to Example 3:
Since the limit of the numerator
$$\lim_{{x \to 1}} (x^2 - 1) = 0$$
and that of the denominator
$$\lim_{{x \to 1}} (x - 1) = 0$$
are both equal to zero, we can use L'Hopital's rule to calculate the limit
$$\lim_{{x \to 1}} \dfrac{{x^2 - 1}}{{x - 1}} = \lim_{{x \to 1}} \dfrac{\dfrac{{d(x^2 - 1)}}{{dx}}}{ \dfrac{{d(x - 1)}}{{dx}}}$$
$$= \lim_{{x \to 1}} \dfrac{{2x}}{{1}} = \dfrac{{2(1)}}{{1}} = 2$$
Note that the same limit may be calculated by first factoring as follows
$$\lim_{{x \to 1}} \dfrac{{x^2 - 1}}{{x - 1}} = \lim_{{x \to 1}} \dfrac{{(x - 1)(x + 1)}}{{x - 1}}$$
$$= \lim_{{x \to 1}} (x + 1) = 2$$

### Example 4

Find the limit $$\lim_{{x \to 2}} \dfrac{{\ln(x - 1)}}{{x - 2}}$$

Solution to Example 4:
Limit of numerator
$$\lim_{{x \to 2}} \ln(x - 1) = 0$$
Limit of denominator
$$\lim_{{x \to 2}} (x - 1) = 0$$
Both limits are equal to zero, L'Hopital's rule may be used
$$\lim_{{x \to 2}} \dfrac{{\ln(x - 1)}}{{x - 2}} = \lim_{{x \to 2}} \dfrac{\dfrac{{d(\ln(x - 1))}}{{dx}}} { \dfrac{{d(x - 2)}}{{dx}} }$$
$$= \lim_{{x \to 2}} \dfrac{{1/(x-1)}}{{1}} = \dfrac{{1/(2 -1)}}{{1}} = 1$$

### Example 5

Find the limit $$\lim_{{x \to 0}} \dfrac{{1 - \cos x}}{{6x^2}}$$

Solution to Example 5:
Limit of numerator and denominator
$$\lim_{{x \to 0}} (1 - \cos x) = 0$$
$$\lim_{{x \to 0}} 6x^2 = 0$$
L'Hopital's rule may be used
$$\lim_{{x \to 0}} \dfrac{{1 - \cos x}}{{6x^2}} = \lim_{{x \to 0}} \dfrac{{\sin x}}{{12x}}$$
The new limit is also indeterminate $$\dfrac{0}{0}$$ and we may apply L'Hopital's theorem a second time
$$= \lim_{{x \to 0}} \dfrac{{\cos x}}{{12}} = \dfrac{{\cos 0}}{{12}} = \dfrac{1}{12}$$

## Exercises

Find the limits
1. $$\lim_{{x \to 0}} \dfrac{{\sin 4x}}{{\sin 2x}}$$
2. $$\lim_{{x \to 0}} \dfrac{{\tan x}}{{x}}$$
3. $$\lim_{{x \to 1}} \dfrac{{\ln x}}{{3x - 3}}$$
4. $$\lim_{{x \to 0}} \dfrac{{e^x - 1}}{{\sin 2x}}$$

### Solutions to Above Exercises

1. $$2$$
2. $$1$$
3. $$\dfrac{1}{3}$$
4. $$\dfrac{1}{2}$$