Evaluating $ \lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right) $

Question

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

Answer 1

That limit is $-\frac1{3!}$. That's so because, when a power $\sum_{n=0}^\infty a_nx^n$ series has radius of convergence $r$ greater than $0$ (and the radius of convergence of your series is $\infty$), then, if $f(x)=\sum_{n=0}^\infty a_nx^n$ ($|x|<r$), $f$ is a continuous function. In particular,$$a_0=f(0)=\lim_{x\to 0}f(x)=\lim_{x\to0}\sum_{n=0}^\infty a_nx^n.$$

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Here is a more elementary approach. For each real number $x$ such that $|x|<1$,\begin{align}\left|\frac{x^2}{5!}-\frac{x^4}{7!}+\cdots\right|&\leqslant\frac{|x|^2}{5!}+\frac{|x|^4}{7!}+\cdots\\&\leqslant\frac{|x|^2}{120}\left(1+|x|^2+|x|^4+\cdots\right)\\&=\frac{|x|^2}{120\left(1-|x|^2\right)}\end{align}And so, since $\lim_{x\to0}\frac{|x|^2}{120\left(1-|x|^2\right)}=0$,$$\lim_{x\to0}-\frac1{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\cdots=-\frac1{3!}+0=-\frac1{3!}.$$

Answer 2

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 $$ \lim_{x\to 0}\frac{1}{x^3}\left(\sin x-x\right)=-\frac{1}{3!} $$