# Thread: Help with L'Hospital's Rule question

1. ## Help with L'Hospital's Rule question

Hey guys,

I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.

Here's the question:

Evaluate the limit.

This is what I have gotten so far:

lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)

Took the deriv. again because I get 0/0
lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.

Any help's appreciated. Thanks!

2. Originally Posted by dark-ryder341
Hey guys,

I was just wondering if you could help me out with a question. I think I've wittled it down enough but I'm having troubles simplifying it and I'm not sure if I've done one step right.

Here's the question:

Evaluate the limit.

This is what I have gotten so far:

lim x --> 0 = sin(8x) / -tan(4x)^-2 (because, 1/1/cot(4x) = tan(4x)^-1, and the derivative of that is what I have written)

Took the deriv. again because I get 0/0
lim x --> 0 = 8cos(8x) / -4sec^2(4x)^-2 << this is the step I'm not sure about, with sec. I don't think that's the right derivative.

Any help's appreciated. Thanks!
$\lim_{x \to 0}\cot{(4x)}\sin{(8x)} = \lim_{x \to 0}\frac{\sin{(8x)}}{\tan{(4x)}}$

Since this tends to the indeterminate $\frac{0}{0}$, we can use L'Hospital.

$= \lim_{x \to 0}\frac{\frac{d}{dx}[\sin{(8x)}]}{\frac{d}{dx}[\tan{(4x)}]}$

$= \lim_{x \to 0}\frac{8\cos{(8x)}}{4\sec^2{(4x)}}$

$= \lim_{x \to 0}2\cos{(4x)}\cos^2{(8x)}$

$= 2\cos{0}\cos^2{0}$

$= 2$.

3. another way : just simplify the term $\sin(8x)\cot(4x)$.
you'll get the same result without using the L'hospital rule.