I want to calculate the limit of $$\lim\limits_{x \to 0} \frac{1-\cosh(x^5)}{x^6\sin^2(x^2)}$$ I know it is $-\infty$, but I want to know how to get there by hand. I have tried L'Hospital, which did not work out and I have tried Taylor expansion:
numerator: $1-\cosh(x^5)=-\dfrac{x^{10}}{2}-\dfrac{x^{20}}{24}+O(x^{26})$
enumerator: $x^6\sin^2(x^2)=x^{10}-\dfrac{x^{14}}{3}+O(x^{16})$
If taylor expansion is the right way to solve it, I don't know how to continue...
$\lim\limits_{x\rightarrow0}\frac{1-\cosh{x}}{x^2}=\lim\limits_{x\rightarrow0}\frac{1-\frac{a^x+e^{-x}}{2}}{x^2}=\lim\limits_{x\rightarrow0}\frac{-(e^x-1)^2}{2e^xx^2}=-\frac{1}{2}$ and $\lim\limits_{x\rightarrow0}\frac{\sin{x}}{x}=1$.
Thus, $$\lim\limits_{x \to 0} \frac{1-\cosh(x^5)}{x^6\sin^2(x^2)}=-\frac{1}{2}\lim\limits_{x \to 0}\frac{x^{10}}{x^{10}}=-\frac{1}{2}$$