# l'Hospital's rule and limits.

• Dec 3rd 2011, 07:15 PM
Pupil
l'Hospital's rule and limits.
$\displaystyle \lim_{x \to \0}\frac{(3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}}}{x}$, which gives the indeterminate form $\displaystyle \frac{0}{0}$. Using l'Hospital' rule gives:
$\displaystyle \lim_{x \to \0}[\frac{4}{7}(3+x)^{-\frac{3}{7}} - \frac{4}{7}(3+x)^{-\frac{3}{7}}] = 0$

But, the book gives a different answer and I confirmed their answer by using a different method. What I want to know exactly is why l'Hospital's rule failed to work for this limit.
• Dec 3rd 2011, 07:37 PM
alexmahone
Re: l'Hospital's rule and limits.
Quote:

Originally Posted by Pupil
$\displaystyle \lim_{x \to \0}\frac{(3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}}}{x}$, which gives the indeterminate form $\displaystyle \frac{0}{0}$. Using l'Hospital' rule gives:
$\displaystyle \lim_{x \to \0}[\frac{4}{7}(3+x)^{-\frac{3}{7}} - \frac{4}{7}(3+x)^{-\frac{3}{7}}] = 0$

But, the book gives a different answer and I confirmed their answer by using a different method. What I want to know exactly is why l'Hospital's rule failed to work for this limit.

The derivative of $\displaystyle (3+x)^{\frac{4}{7}} - (3-x)^{\frac{4}{7}}$ is $\displaystyle \frac{4}{7}(3+x)^{-\frac{3}{7}}+\frac{4}{7}(3-x)^{-\frac{3}{7}}$.