Bernoulli-L'Hôpital's Rule

analysis calculus

Definition

L'Hospital's Rule

Let $f$ and $g$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Let $x_0\in [a,b]$ such that:

$$\underset{x\to x_0}{\lim }f(x)=\underset{x\to x_0}{\lim }g(x)=0\text{or}\pm \infty$$

If the limit of the ratio of their derivatives exists, then:

$$\underset{x\to x_0}{\lim }\frac{f(x)}{g(x)}=\underset{x\to x_0}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$$

providing $g^{\prime }(x)\ne 0$ in a neighbourhood of $x_0$.

Examples

${\lim }_{x\to 0}\frac{\sin (x)}{x}$

The fraction results in an indeterminate form $\frac{0}{0}$. Applying L’Hospital’s rule:

$$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{1}=\cos (0)=1$$

${\lim }_{x\to 0}\frac{1-\cos (x)}{x^2}$

This results in $\frac{0}{0}$. Applying the rule twice:

$$\underset{x\to 0}{\lim }\frac{1-\cos (x)}{x^2}=\underset{x\to 0}{\lim }\frac{\sin (x)}{2x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{2}=\frac{1}{2}$$

${\lim }_{x\to \infty }\frac{x^n}{e^x}$

This results in the indeterminate form $\frac{\infty }{\infty }$. Applying the rule $n$ times:

$$\underset{x\to \infty }{\lim }\frac{x^n}{e^x}=\underset{x\to \infty }{\lim }\frac{n{x}^{n-1}}{e^x}=\cdots =\underset{x\to \infty }{\lim }\frac{n!}{e^x}=0$$