# L'Hopital's Rule

• Oct 19th 2008, 05:03 PM
Truthbetold
L'Hopital's Rule
$\displaystyle f(x)= \frac{1- cos (x)}{x + x^2}$

Find the error in the following incorrect application of L'Hopital's rule.

$\displaystyle \frac{f'(x)}{g'(x)}= \frac{sin (x)}{1 + 2x}$
$\displaystyle \frac{f''(x)}{g''(x)}= \frac{cos (x)}{2}$

Therefore, the lim x--> 0 = $\displaystyle \frac{1}{2}$.

I'm not sure what the problem is. My guess is that it has something to do with ignoring the 0 (that came from the derivative of 1) on the numerator of the 2nd derivative.

I know the value of 0 plugged in is correct. There didn't appear any faulty differentiation.

Another thought is that I should use a trig rule.
1 - cos (x) looks like it should be replaced by something.
• Oct 19th 2008, 05:13 PM
Krizalid
$\displaystyle \frac{1-\cos x}{x+x^2}\to\frac00$ as $\displaystyle x\to0,$ and $\displaystyle \frac{\sin x}{1+2x}\to0$ as $\displaystyle x\to0,$ hence, there's no reason to differentiate again since there's no indeterminate form.