« Limits of Trigonometric Functions

L'Hospital's Rule for $\frac{0}{0}$

Suppose that $\lim f(x) = \lim g(x) = 0$. Then

If $\lim \frac{f'(x)}{g'(x)} = L$, then $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} = L$.

If $\lim \frac{f'(x)}{g'(x)}$ tends to $\infty$ in the limit, then so does $\frac{f(x)}{g(x)}$.

Here is a case of $\frac{0}{0}$.

Example 1:

$$ \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = \mathop {\lim }\limits_{x \to 0} \frac{(\sin x)'}{x'} = \mathop {\lim }\limits_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1 $$

Example 2:

$$ \mathop {\lim }\limits_{x \to 1} \frac{{\ln x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(\ln x)'}}{{(x - 1)'}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{x}}}{1} = \mathop {\lim }\limits_{x \to 1} \frac{1}{x} = 1 $$

Example 3:

$$ \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{({e^x} - 1)'}}{{({x^2})'}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{2x}} = \infty $$

Example 4:

Sometimes it is necessary to use L'Hospital's Rule several times in the same problem.

$$ \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{2} = \frac{1}{2} $$

Example 5:

Here is a more elaborate example involving the indeterminate form $\frac{0}{0}$. Applying the rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying L'Hospital's rule three times:

$$ \begin{aligned} &\mathop {\lim }\limits_{x \to 0} \frac{{2\sin x - \sin 2x}}{{x - \sin x}} = \\ &= \mathop {\lim }\limits_{x \to 0} \frac{{2\cos x - 2\cos 2x}}{{1 - \cos x}} = \\ &= \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin x + 4\sin 2x}}{{\sin x}} = \\ &= \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\cos x + 8\cos 2x}}{{\cos x}} = \\ &= \frac{{ - 2\cos 0 + 8\cos 0}}{{\cos 0}} = \\ &= 6 \end{aligned} $$

L'Hospital's Rule for $\frac{\infty }{\infty }$

Suppose $\lim f(x) = \lim g(x) = \infty$. Then

If $\lim \frac{f'(x)}{g'(x)} = L$, then $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} = L$.

If $\lim \frac{f'(x)}{g'(x)}$ tends to $+ \infty$ or $- \infty$ in the limit, then so does $\frac{f(x)}{g(x)}$.

Here is a case of $\frac{\infty }{\infty }$

Example 6:

$$ \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{{\ln x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1/(2\sqrt x )}}{{1/x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x }}{2} = \infty $$

Example 7:

$$ \mathop {\lim }\limits_{x \to \infty } {x^n}{e^{ - x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n{x^{n - 1}}}}{{{e^x}}} = n\mathop {\lim }\limits_{x \to \infty } \frac{{{x^{n - 1}}}}{{{e^x}}} $$

Iterate the above until the exponent is 0. Then one sees that the limit is 0.

Example 8:

This one also involves $\frac{\infty }{\infty }$:

$$ \mathop {\lim }\limits_{x \to 0 + } (x\ln x) = \mathop {\lim }\limits_{x \to 0 + } \frac{{\ln x}}{{1/x}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{1/x}}{{ - 1/{x^2}}} = \mathop {\lim }\limits_{x \to 0 + } - x = 0 $$

Basic indeterminate forms (all others reduce to these):

$$\frac{0}{0} \text{ and } \frac{\infty }{\infty }$$

Other indeterminate forms:

$$\infty ^0 \ \ \ 1^\infty \ \ \ 0 \cdot \infty \ \ \ 0^0 \ \ \ \infty - \infty$$

Example 9:

To handle a case of $\infty - \infty$, the difference of two functions is converted to a quotient:

$$ \begin{aligned} &\mathop {\lim }\limits_{x \to \infty } x - \sqrt {{x^2} - x} = \mathop {\lim }\limits_{x \to \infty } \frac{{(x + \sqrt {{x^2} - x} )(x - \sqrt {{x^2} - x} )}}{{x + \sqrt {{x^2} - x} }} = \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - ({x^2} - x)}}{{x + \sqrt {{x^2} - x} }} = \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{x}{{x + \sqrt {{x^2} - x} }} = \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + \frac{{2x - 1}}{{2\sqrt {{x^2} - x} }}}} = \frac{1}{{1 + 1}} = \frac{1}{2} \end{aligned} $$

Example 10:

To handle a case of $0^0$, the difference of two functions is converted to a quotient:

$$\mathop {\lim }\limits_{x \to 0} {x^x} = \mathop {\lim }\limits_{x \to 0} {e^{\ln ({x^x})}} = \mathop {\lim }\limits_{x \to 0} {e^{x\ln x}} = \mathop {\lim }\limits_{x \to 0} {e^0} = 1$$

When NOT to use L'Hospital's rule

$$\mathop {\lim }\limits_{x \to 0} \frac{\cos x}{x} \ne \mathop {\lim }\limits_{x \to 0} \frac{(\cos x)'}{x'}$$

because $\mathop {\lim }\limits_{x \to 0} \cos x = 1 \color{red}{\ne} 0$