Computing $\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$

Question

So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$ The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas WolframAlpha used L'Hopital's Rule in solving it. My first instinct to solving the question when I read it was like this- $$ \lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}} $$ $$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \frac{1}{x^2} - \frac{x}{x^3} $$ $$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{x}{x^3} $$ $$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{1}{x^2} $$ $$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} $$ $$ \implies 1 \times 0 $$ $$ \implies 0 $$

I do not understand, where I am going wrong with this procedure. I would really appreciate it if someone can answer. Thank you.

For reference here is the solution given in Cengage.

Answer 1

The limit $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right)$$ is of the indeterminate form $\infty - \infty$ so $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right) \neq \lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2}\right) - \lim_{x\to 0}\left(\frac 1{x^2}\right)$$ because the RHS makes no sense.

Answer 2

Proper evaluation ultimately rests on the result

$\underset{x\to0}{\lim}\dfrac{\sin(x)}{x}=1,$

as methods based on differentiation (including Taylor seties) ultimately require this fact to establish the needed derivate of the sine function.

Below is a method that properly uses the aboceresilt without introducing infinite quantities or requiring differentiation:

Let

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)-x}{x^3}$

and multiply in a factor of $\cos(x)$ whose limit is just $1$. Thus

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)\cos(x)-x\cos(x)}{x^3}$

We apply thedoue angle sine formula and reintroduce the limit on the right side of the equation:

$L=\underset{x\to0}{\lim}\dfrac{(1/2)\sin(2x)-x\cos(x)}{x^3}$

$=\underset{x\to0}{\lim}\dfrac{(1/2)(\sin(2x)-2x)-x(-1\cos(x))}{x^3}$

$=\underset{x\to0}{\lim}\dfrac{(4)(\sin(2x)-2x)}{(2x)^3}-\underset{x\to0}{\lim}\dfrac{-1\cos(x)}{x^2}$

$=4L-\underset{x\to0}{\lim}\dfrac{1-\cos(x)}{x^2}$

Thus

$3L=-\underset{x\to0}{\lim}\dfrac{1-\cos(x)}{x^2}$

$=-(1/2)\underset{x\to0}{\lim}\dfrac{\sin^2(x/2)}{x^2}=-1/2$

where we have use the half-angle sine formula and then the $\sin(x)/x$ limit. Thereby

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)-x}{x^3}=-1/6.$

Answer 3

Assuming the limit exists, let $$l=\lim_{x \to 0}\frac{\sin x-x}{x^3} $$ $$ l=\lim_{x \to 0}\frac{\sin x-x}{x^3}\overset{x=2t}{=}\lim_{t \to 0}\frac{\sin 2t-2t}{8t^3}=\lim_{x \to 0}\frac{2\sin x \cos x-2x}{8x^3} $$ Hence $$4l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3} $$
Now consider $$3l=4l-l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3}-\lim_{x \to 0}\frac{\sin x-x}{x^3}=\lim_{x \to 0}\frac{\sin x \cos x-\sin x}{x^3}\\ 3l=\lim_{x \to 0}\frac{\ \sin x}{x} \lim_{x \to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2} $$ Finally $$ \boxed{l=\lim_{x \to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}}$$