I am trying to find the one sided limits numerically for this equation:
$$\lim_{x\to0_\pm} \frac{x−\sin\left|x\right|}{x^3}$$
I know the answer for $\lim_{x\to0_{-}}$ is $\infty$, and $\lim_{x\to0_{+}}$ is $\frac16$ but I am having the hardest time arriving at these conclusions.
Thank you all for the help.
If the question says numerically, that means that it wants you to do something like this:
If $f(x) = \frac{x - \sin |x|}{x^3}$, then the numerical limit would be calculated like this:
The limit as $x$ approaches the left: $f(x-h)$ for very small $h$. (i.e. Make an $h$ like $.0001$ and plug it into a calculator)
Limit as $x$ approaches from the right: $f(x+h)$ for every small $h$.
In this answer, it is shown that $$ \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{1} $$ Now, it is pretty easy to show that $$ \begin{align} \lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3} &=\lim_{x\to0}\frac{\tan(x)(1-\cos(x))}{x^3}\\ &=\lim_{x\to0}\frac{\tan(x)\sin^2(x)}{x^3(1+\cos(x))}\\ &=\lim_{x\to0}\frac{\tan(x)}{x}\left(\frac{\sin(x)}{x}\right)^2\frac1{1+\cos(x)}\\ &=\frac12\tag{2} \end{align} $$ Subtract the reciprocal of $(1)$ from $1$ to get $$ \lim_{x\to0}\frac{\tan(x)-\sin(x)}{x-\sin(x)}=3\tag{3} $$ Divide $(2)$ by $(3)$ to get $$ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag{4} $$ This gives the right side limit. The left side limit is much simpler.
The Taylor formula $f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}2+f'''(0)\frac{x^3}6+o(x^3)$ for $\sin x$ provides $\sin x=x-\displaystyle\frac{x^3}6+o(x^3)$. Therefore, $\displaystyle\frac{x-\sin|x|}{x^3}=\frac16+o(1)$ if $x>0$ and $\displaystyle\frac{x-\sin|x|}{x^3}=\frac2{x^2}-\frac16+o(1)$ if $x<0$. (Here, $o\big(f(x)\big)$ denotes a function $g(x)$ such that $\lim\limits_{x\to0}\displaystyle\frac{g(x)}{f(x)}=0$.) You can also prove these limits using l'Hôpital's rule in place of the Taylor formula.
