How do you find the limit using L'Hospital's Rule?
$$\lim_{x \rightarrow 0}\frac{ x-sinx }{ \tan^3x }$$

Question

How do you find the limit using L'Hospital's Rule?
$$\lim_{x \rightarrow 0}\frac{ x-sinx }{ \tan^3x }$$

misty1212 · Answer

hi
take the derivative separately top and bottom, try again

misty1212 · Answer

it won't work the first time, have to do it again

misty1212 · Answer

first attempt gives $$\frac{1-\cos(x)}{3\tan^2(x)\sec^2(x)}$$ still no good
have to do it one more time

hpfan101 · Answer

Oh okay, I'm going to show you what I got the second time to make sure I'm doing the derivative right

hpfan101 · Answer

$$\frac{ sinx }{ 6tanx \times \sec^2x+3\tan^2x \times 2secx \times secx \times tanx }$$

misty1212 · Answer

looks right,maybe there is a better way, because you still get 0/0

hpfan101 · Answer

So would I take the derivative again? Or do I try to simplify?

misty1212 · Answer

maybe we can do something else before that step

hpfan101 · Answer

I was thinking about rewriting the 6tanx as (6)(sinx/cosx)? Not sure if that'll help

misty1212 · Answer

ack i dont see a good way to change this to make it easier

misty1212 · Answer

guess you have to go again

zarkon · Answer

$$\lim_{x\to0}\frac{1-\cos(x)}{3\tan^2(x)\sec^2(x)}=\lim_{x\to0}\frac{1-\cos(x)}{3\tan^2(x)}\lim_{x\to0}\frac{1}{\sec^2(x)}$$

misty1212 · Answer

guess that is a good way!

hpfan101 · Answer

Oh, so two different limits? Didn't know we can do that