# Math Help - Finding limits

1. ## Finding limits

If I want to find the $\lim_{x\rightarrow 0}\frac {xcosx -sin(x)}{x^2}$, using L'Hopital's rule, also knowing $\lim_{x\rightarrow 0}\frac{sinx}{x}$ is equal to 1, how many times should I differentiate?

i differentiated it once, and my answer was that the limit = -1

Is this correct?

2. I do not find the same result as you, apart from the fact that I needed only to differentiate once !

3. $\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ differentiate once:

$\lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}$ so

$\lim_{x\rightarrow 0}\frac{-xsinx}{2x}$ eliminate $x$ (which you didn´t) and you have that the limit is 0

4. Originally Posted by Inti
$\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ differentiate once:

$\lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}$ so

$\lim_{x\rightarrow 0}\frac{-xsinx}{2x}$ eliminate $x$ (which you didn´t) and you have that the limit is 0
Hey, thanks for the quick reply.

I can't see how $\frac{-xsinx}{2x}$ is the derivative of [tex]\frac{cosx-xsinx-cosx}{2x}[/math. Could you please explain?

Thanks again.

5. I just cancelled the opposite cosines, I wasn´t differentiating...

6. Originally Posted by Inti
I just cancelled the opposite cosines, I wasn´t differentiating...
Sorry Inti, but I still can't see what you're doing...

$\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$

Then you differentiate it, right? or what did you do to get the following?

$\lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}$

I can see how you can cancel $cosx$ after differentiating...

See, the problem is now that my derivative of $\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ is $\lim_{x\rightarrow 0}\frac{-x^2sinx+2xcosx-2sinx}{x^3}$

7. Originally Posted by tsal15
Sorry Inti, but I still can't see what you're doing...

$\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$

Then you differentiate it, right? or what did you do to get the following?

$\lim_{x\rightarrow 0}\frac{cosx-xsinx-cosx}{2x}$

I can see how you can cancel $cosx$ after differentiating...

See, the problem is now that my derivative of $\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ is $\lim_{x\rightarrow 0}\frac{-x^2sinx+2xcosx-2sinx}{x^3}$
OK tsal15 just a minute
Differentiate $\frac{xcosx-sinx}{x^2}$ is $\frac{-x^2sinx-2xcosx+2sinx}{x^3}$ (sign mistake I think)

Anyway it does not matter because this is not what L'Hopital's rule says
It says
$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}\:=\:\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}$

You do not need to differentiate $\frac{f(x)}{g(x)}$

Therefore
$\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ is $\lim_{x\rightarrow 0}\frac{-xsinx}{2x}$

8. **** yeah!!! I was going to ask exactly that: if he/she was differentiating THE ENTIRE FUNCTION instead of the dividend and the divisor separately...

9. Originally Posted by Inti
**** yeah!!! I was going to ask exactly that: if he/she was differentiating THE ENTIRE FUNCTION instead of the dividend and the divisor separately...
I get you now...So, the limit is equal to ZERO?

10. Originally Posted by running-gag
OK tsal15 just a minute
Differentiate $\frac{xcosx-sinx}{x^2}$ is $\frac{-x^2sinx-2xcosx+2sinx}{x^3}$ (sign mistake I think)

Anyway it does not matter because this is not what L'Hopital's rule says
It says
$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}\:=\:\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}$

You do not need to differentiate $\frac{f(x)}{g(x)}$

Therefore
$\lim_{x\rightarrow 0}\frac{xcosx-sinx}{x^2}$ is $\lim_{x\rightarrow 0}\frac{-xsinx}{2x}$
OF COURSE!!! DOH!!! thanks, running-gag. so, the limit is equal to 0 (zero)?