How is limit of the function $(x-\sin x)/(x^3)$ where $x\to 0$, equal to $1/6$?

Question

I am basically confused as to why this is equal to $\frac{1}{6}$?

I know about the sine series using the Maclaurin series, also about the sine template which is if $f(x) = \frac{\sin x}{x}$ where $x\to 0$ then $f(x)=1$.

Please help me with the proof of the aforementioned question.

Thank you.

https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion — lab bhattacharjee, May 05 '21 at 07:26

score 9 · Answer 1 · answered May 05 '21 at 07:26

9

Ponder the limit of

$$\frac{x-\left(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots\right)}{x^3}=\dfrac1{3!}+\dfrac{x^2}{5!}-\dfrac{x^4}{7!}+\cdots$$

answered May 05 '21 at 07:26

score 2 · Answer 2 · answered May 05 '21 at 07:25

You could show it by using l'Hospitals rule twice. This is a theorem that states (simplified for your case): If $f$ and $g$ are twice differentiable functions, such that $f(0)=g(0)=f'(0)=g'(0)=0$ and if there is an intervall around $0$ such that $\frac fg$ and $\frac{f'}{g'}$ are well defined, then:

$$\lim\limits_{x\rightarrow 0} \frac{f(x)}{g(x)}=\lim\limits_{x\rightarrow 0} \frac{f'(x)}{g'(x)}=\lim\limits_{x\rightarrow 0} \frac{f''(x)}{g''(x)}$$, which will yield your answer directly.

How is limit of the function $(x-\sin x)/(x^3)$ where $x\to 0$, equal to $1/6$?

2 Answers2