How do I evaluate the limit $$ \lim_{x \to 0}\frac{e^{x^2-x} -1 + x - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$
depending on $\alpha, $using Taylor series? I know I consider
$e^t = 1 +t +\frac{t^2}{2} + o(t^2)$
$ \log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + o(t^3)$
$\cosh(t) = 1+ \frac{t^2}{2} + \frac{t^4}{24} + o(t^4)$
$\sin(t) = t - \frac{t^3}{6} + \frac{t^5}{120} + o(t^5)$
and substitute. I know why it is possibile and to justify this method. Though, I am not able to solve this limit... Thanks!
Note that
$$\begin{cases}x^3 \cdot x \sin (1/x^2)\to 0 \implies x^4\sin (1/x^2)=o(x^3)\\\\ x^3 \cdot x\log x\to 0\implies x^4\log x=o(x^3)\\\\ e^{x^2-x} -1 =x^2-x+\frac{(x^2-x)^2}{2}+o(x^2)=-x+\frac32x^2+o(x^2)\\\\ \cosh 2x=1+2x^2+o(x^2)\end{cases}$$
thus
$$\frac{e^{x^2-x} -1 -\alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)} =\frac{-x+\frac32x^2+o(x^2) - \alpha x^2 + o(x^3) }{1+2x^2+o(x^2) -1 + o(x^3)} =\frac{-x+\left(\frac32-\alpha \right)x^2 +o(x^2) }{2x^2+o(x^2)}=\frac{-\frac1x+\left(\frac32-\alpha \right) +o(1) }{2+o(1)}\to\pm\infty $$
therefore $$\lim_{x \to 0}\frac{e^{x^2-x} -1 - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$
does not exist.