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.

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

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

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

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

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

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

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

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

$\displaystyle L = 1$

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