L' Hospital's Rule

analysis

Definition

L' Hospital's Rule

Let the functions $f$ and $g$ be continuous on $[a,b]$ and differentiable in the interior $(a,b)$. Furthermore, let $x_0\in [a,b]$ and let $f(x_0)=g(x_0)=0$. Also, it is assumed that the limit ${\lim }_{x\to x_0}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists (or the one-sided limit, if $x_0=a$ or $x_0=b$). Then the following holds:

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

Examples:

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

$$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=\frac{{\lim }_{x\to 0}\sin (x)}{{\lim }_{x\to 0}x}\stackrel{\sin (x)\text{ continuous}}{=}\frac{\sin ({\lim }_{x\to 0}x)}{{\lim }_{x\to 0}x}=\frac{0}{0}$$

The fraction $\frac{0}{0}$ is indeterminate. Applying L’Hospital’s rule:

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

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

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

The fraction $\frac{0}{0}$ is indeterminate. Applying L’Hospital’s rule:

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

Again, the fraction $\frac{0}{0}$ is indeterminate. Applying L’Hospital’s rule:

$$\underset{x\to 0}{\lim }\frac{\sin x}{2x}=\underset{x\to 0}{\lim }\frac{\left(\frac{\partial }{\partial x}\sin (x)\right)}{\frac{\partial }{\partial x}2x}=\underset{x\to 0}{\lim }\frac{\cos (x)}{2}=\frac{{\lim }_{x\to 0}\cos (x)}{2}=\frac{\cos ({\lim }_{x\to 0}x)}{2}=\frac{1}{2}$$

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

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

The fraction $\frac{\infty }{\infty }$ is indeterminate. Applying L’Hospital’s rule:

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