1. ## l'hopitals rule problem

can someone solve

lim (sin4x)/(x^2+8x)
x approaches 0

2. Originally Posted by cummings15
can someone solve

lim (sin4x)/(x^2+8x)
x approaches 0
$\displaystyle \lim_{x\to 0}\frac{\sin (4x)}{x^2+8}=\lim_{x\to 0}\frac{\cos(4x)\cdot 4}{2x+8}=\frac{4}{8}=\frac{1}{2}$

3. A little more basic method:

cummings15, rewrite it as $\displaystyle \frac{sin(4x)}{x^2+ 8x}= \frac{sin(4x)}{x(x+ 8)}$$\displaystyle = 4\frac{sin(4x)}{4x}\frac{1}{x+ 8}$.

Now, if you know the $\displaystyle \lim_{\theta\to 0}\frac{sin(\theta)}{\theta}$, this is easy.

4. ## :(

wats actually a l'hopital rule

5. Yes, I just noticed the title of the thread- but I still think the more fundamental method is better.

6. ## yea got it

l'hospital is a rule to differentiate numerator and denominator separately and then substitute the limit

can we use it for 1/0 cases or infinity/0 cases??