# Thread: [SOLVED] Finding a limit using l'Hopital's Rule

1. ## [SOLVED] Finding a limit using l'Hopital's Rule

I thought I had this one, but my answer does not agree with what I got when I checked my answer against a calculator graph of the equation. I got lim = 1, the graph shows lim = 0.

$L = \lim _{x\rightarrow0}\left(\cot{x} - \frac{1}{x}\right)$

$L = \lim_{x\rightarrow0}\left(\frac{1}{\tan{x}} - \frac{1}{x}\right)$

$\ln{L} = \ln{\lim_{x\rightarrow0}\left(\frac{1}{\tan{x}} - \frac{1}{x}\right)}$

$\ln{L} = \lim_{x\rightarrow0}\ln\left(\frac{\frac{1}{\tan{x }}}{\frac{1}{x}}\right)$

$\ln{L} = \lim_{x\rightarrow0}\ln\left(\frac{x}{\tan{x}}\rig ht)$

$\ln{L} = \ln\lim_{x\rightarrow0}\left(\frac{x}{\tan{x}}\rig ht) \ \mbox{Here is the 0/0 form for l'Hopital's rule.}$

$\ln{L} = \ln\lim_{x\rightarrow0}\left(\frac{1}{\sec{x}^2}\r ight)$

$\ln{L} = \ln{1}$

$L = 1$

So, where did I go wrong? I have looked over this a dozen times and can not see the problem. Thanks.

2. $ln\left(\frac{1}{tan(x)}-\frac{1}{x}\right)\neq ln\left(\frac{\frac{1}{tan(x)}}{\frac{1}{x}}\right )$

You are thinking of $ln\left(\frac{\frac{1}{tan(x)}}{\frac{1}{x}}\right )=ln(cot(x))-ln(\frac{1}{x})$

This is a common mix up.

3. $\lim_{x\to 0}\frac{1}{\tan x}-\frac{1}{x}=\lim_{x\to 0}\frac{x-\tan x}{x\tan x}\overbrace{=}^{\text{L'H}}\lim_{x\to 0}\frac{1-\sec^2 x}{\tan x+x\sec^2 x}$ $\overbrace{=}^{\text{L'H}}\lim_{x\to 0}\frac{2\sec^2 x\tan x}{2\sec^2 x+2x\sec ^2 x \tan x}\to 0$

4. Thanks for the help! I will have to watch the ln properties -- I am a little rusty. Abu-Khalil's way works better, anyway. I wish I gone that route to begin with -- it would have saved me a lot of time and frustration.