Hi.
Evaluate.
$\displaystyle \lim_{\ x\to 0}\frac{\log{cosx}-xsinx}{2e^{x^2}-cos2x-1}$
Look, this is kinda' news to me but it seems to fall out right.. So if we can obtain the Taylor series:
$\displaystyle \log(\cos(x))-x\sin(x)=-3/2 x^2+x^4/10+\cdots$
$\displaystyle 2e^{x^2}-\cos(2x)-1=4x^2+x^4/3+\cdots$
and then use long division:
$\displaystyle
\begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, } \vspace*{0.12cm} \\
\cline{2-3}
\multicolumn{1}{r}{4x^2+x^4/3+\cdots \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} -3/2 x^2+x^4/12+\cdots} \\
& \multicolumn{2}{l}{\phantom{xxxxxxxxxxxxxxxxxxxxxx }}
\end{array}
$
note the first term is -3/8 and I assume the remainder will have powers of x so that as x->0, the limit is -3/8 which is the limit given by Mathematica.
I'm not entirely sure about the validity of this method with the other one since $\displaystyle \sqrt{x}$ has a branch-point at the origin.
Why not skeeter? They're both analytic at zero so it's easy to compute the first few terms of the Taylor series for each just be taking a few derivatives at x=0. I guess all this is ok. Never solved a limit this way before. Maybe some issues of uniform convergence is required for it to be valid though.