L'Hopital's rule.

Theorem 6.14.1   Let $ f$ and $ g$ be two functions verifying the following conditions:
  1. The functions $ f$ and $ g$ are both differentiable on a pointed neighborhood of $ x_0$ ;
  2. $ \underset {x\longrightarrow x_0}{\text{lim}} g'(x) \neq 0$ ;
  3. $ \underset {x\longrightarrow x_0}{\text{lim}} f(x)= \underset {x\longrightarrow x_0}{\text{lim}} g(x)=0$ . Then $ \underset{x \longrightarrow x_0}{\text{lim}} \frac {f(x)}{g(x)} =
\underset {x\longrightarrow x_0}{\text{lim}} \frac {f'(x)}{g'(x)}$ .

Example 6.14.2   We can verify here some limits from Table 4.1:

$\displaystyle \underset{x \longrightarrow 0}{\text{lim}} \frac {\sin x}{x} = \left[ \frac {\cos x}{1} \right]_{x=0}=1.$    


$\displaystyle \underset{x \longrightarrow 0}{\text{lim}} \frac {e^x-1}{x}=\left[ \frac {e^x}{1} \right]_{x=0}=1.$    

If $ f'(x_0)=g'(x_0)=0$ , we can iterate the processus described in Theorem  14.1. For example:

$\displaystyle \underset{x \longrightarrow 0}{\text{lim}} \frac {\sin^2 x}{e^x -...
...tal}}{=} \underset{x \longrightarrow 0}{\text{lim}} \frac {2 \cos 2x}{e^x} = 2.$    

Example 6.14.3  

$\displaystyle \underset{x \longrightarrow 0}{\text{lim}} \left( \frac 1x - \fra...
... = \underset{x \longrightarrow 0}{\text{lim}} \frac {e^x}{xe^x+2e^x} =\frac 12.$    


The result in Theorem 14.1 is unchanged if we replace $ \underset {x\longrightarrow x_0}{\text{lim}} f(x)= \underset {x\longrightarrow x_0}{\text{lim}} g(x)=0$ by $ \underset {x\longrightarrow x_0}{\text{lim}} f(x)= \infty$ and $ \underset {x\longrightarrow x_0}{\text{lim}} g(x)= \infty$ .

Example 6.14.4  

$\displaystyle \underset{x \longrightarrow +\infty }{\text{lim}} \frac { \ln x}{x} = \underset{x \longrightarrow +\infty }{\text{lim}} \frac { 1/x}{1} = 0.$    

Remark 6.14.5   L'Hopital's rule can be proven, and generalized, using Taylor developments. This material is explained in next chapter.

Noah Dana-Picard 2007-12-28