What is the limit of :
$$ \lim_{x\to 0} \frac{\sin(ax) - \ln(1-2x)}{e^{ax}-1-2x-2x^{2}}$$
I did this with Maclaurin, because my exam is about solving these with MacLaurin.
Gave $$\lim_{x\to 0} \frac{ax-\frac {(ax)^{2}}{3!} +2x +4x^{2}}{1+ax+(ax)^{2} -1-2x-2x^2} = \frac {a+2}{a-2}$$
The answer should be: $a = 2$, limit is $1/2$, $ a \neq 2$, limit is $1$
$\displaystyle\lim_{x\to 0}\frac{\sin(ax)-\ln(1-2x)}{e^{ax}-1-2x-2x^2}=\lim_{x\to 0}\frac{ax-\frac{(ax)^3}{3!}+\cdots-[-2x-\frac{(2x)^2}{2}-\frac{(2x)^3}{3}+\cdots]}{1+ax+\frac{(ax)^2}{2!}+\frac{(ax)^3}{3!}+\cdots-1-2x-2x^2}$
$=\displaystyle\lim_{x\to 0}\frac{(a+2)x+2x^2+(\frac{8}{3}-\frac{a^3}{6})x^3+\cdots}{(a-2)x+(\frac{a^2}{2}-2)x^2+\frac{a^3}{6}x^3+\cdots}=\lim_{x\to 0}\frac{(a+2)+2x+(\frac{8}{3}-\frac{a^3}{6})x^2+\cdots}{(a-2)+(\frac{a^2}{2}-2)x+\frac{a^3}{6}x^2+\cdots}$
$\displaystyle=\begin{cases} \frac{a+2}{a-2} &\mbox {, if } a\ne2 \\ \text{does not exist} &\mbox{, if } a=2\end{cases}$