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

• It looks like $\frac{\sin(x)-x}{x^3}$ – Gribouillis Jul 10 '20 at 8:09
• All the terms will tend to zero apart from the first one, so there is your limit – Henry Lee Jul 10 '20 at 8:26
• @HenryLee That argument does not work for infinite sums. – José Carlos Santos Jul 10 '20 at 9:46
• @JoséCarlosSantos Why will it not? – wesupportthepalace Jul 10 '20 at 14:36
• Take, for instance, for each $n\in\Bbb N$ and each $x\in(0,\infty)$,$$f_n(x)=\begin{cases}nx&\text{ if }x\in\left(0,\frac1n\right)\\1&\text{ otherwise.}\end{cases}$$Now, for each $n\in\Bbb N$, let$$g_n=\begin{cases}f_1&\text{ if }n=1\\f_n-f_{n-1}&\text{ otherwise.}\end{cases}$$Then\begin{align}(\forall x\in(0,\infty):\sum_{n=1}^\infty g_n(x)&=f_1(x)+\bigl(f_2(x)-f_1(x)\bigr)+\bigl(f_3(x)-f_2(x)\bigr)+\cdots\\&=\lim_{n\to\infty}f_n(x)\\&=1\end{align}and therefore $\lim_{x\to0}\sum_{n=1}^\infty g_n(x)=1$. However, for each $n\in\Bbb N$, $\lim_{x\to0}g_n(x)=0$. – José Carlos Santos Jul 10 '20 at 14:52

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 $$f(x)=\sum_{n=0}^\infty a_nx^n$$ ($$|x|), $$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.$$

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 $$\lim_{x\to0}\frac{|x|^2}{120\left(1-|x|^2\right)}=0$$,$$\lim_{x\to0}-\frac1{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\cdots=-\frac1{3!}+0=-\frac1{3!}.$$

• I’m a high school student and unable to understand your answer – dRIFT sPEED Jul 10 '20 at 8:19
• @pRSmHJN1 Executive summary: as long as $f(x)=a_0+a_1x+a_2x^2+\cdots$ converges somewhere away from zero, then $\lim_{x\to0}f(x)=a_0$. – Angina Seng Jul 10 '20 at 8:27
• @pRSmHJN1 I have added a more elementary answer. – José Carlos Santos Jul 10 '20 at 8:28
• +1 for the elementary proof. Either one has to use ideas of power series, uniform convergence, or settle down for a elementary proof suited for specific series, but in no case one should be made to believe that the result is automatic. – Paramanand Singh Jul 10 '20 at 12:22

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