Evaluating $ \lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right) $ This question comes to my mind immediately after asking this question.
I was earlier unknown that limit of sum equal sum of limits only when there are finite terms. Now the problem is then how do I evaluate the following limit which earlier I used to do by applying individual limits.
$$
\lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right)
$$
I’m high school student

Observe that, for $x\ne 0$,
$$ -\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots=\frac{1}{x^3} \left(-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots\right)
=\frac{1}{x^3} \left(\frac{x}{1!}-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots-\frac{x}{1!}\right) = \frac{1}{x^3}\left(\sin x-x\right) \\
$$
Then you may apply L'Hôpital three times and obtain that
$$
\lim_{x\to 0}\frac{1}{x^3}\left(\sin x-x\right)=-\frac{1}{3!}
$$
