Show that:-
$$\lim_{x \rightarrow 0 } \frac{1}{6x^2} - \frac{1}{120x^4} +.... ({-1})^{n+1}\frac{1}{(2n+1)!x^{2n}}$$ for $n\geq1$ where n $\in$ $\mathbb{Z^+}$ is equal to $1$ as $n\to\infty$.
My method was as the form of the limit is $\infty - \infty$ , so i tried first assuming limit exists and taking common denominator for a finite n and tried with lhopital if the limit exists but it was not working , may anyone tell ?
$$S=\sum_{n=1}^{\infty}({-1})^{n+1}\frac{1}{(2n+1)!x^{2n}}$$
$$\frac{e^{x}-1-x}{x}=\sum_{n=1}^{\infty}\frac{x^{n}}{(n+1)!} $$
$$\frac{e^{1/x}-1-\frac1x}{\frac1x}=\sum_{n=1}^{\infty}\frac{1}{(n+1)!x^{n}}$$
$$\frac{e^{1/x}-1-\frac1x}{\frac1{x}}+\frac{e^{1/-x}-1+\frac1x}{\frac1{-x}}=2\sum_{n=1}^{\infty}\frac{1}{(2n+1)!x^{2n}}$$
$$x({e^{1/x}}-{e^{1/-x}}-\frac2x)=2\sum_{n=1}^{\infty}\frac{1}{(2n+1)!x^{2n}}$$
Substituting $ x \to ix$
$$ix({e^{1/ix}}-{e^{1/-ix}}-\frac{2}{ix})=-2\sum_{n=1}^{\infty}\frac{({-1})^{n+1}}{(2n+1)!x^{2n}}$$
$$S=-\frac{ix}2(e^{-i/x}-e^{i/x}+\frac{2i}x)$$
$$\lim_{x\to0}S=1$$
