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

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.

• $\lim_{x\to 0}{1\over x^2}=+\infty.$ Dec 26, 2022 at 8:16
• The limitation $\lim \frac{\sin x}{x^3} != \lim \frac{\sin x}{x}\lim \frac{1}{x^2}$ since the $\lim \frac{\sin x}{x^3}$ does not exist,
– Chia
Dec 26, 2022 at 8:17
• And none of your “implications” are actually implications. You should use the equals sign instead (except that in the very first step, the equality would be false). Dec 26, 2022 at 8:22
• The main problem is that the equality $\lim_{x\to a} (f(x)+g(x))=\lim_{x\to a} f(x) + \lim_{x\to a} g(x)$ is true when both the limits $\lim_{x\to a} f(x)$ and $\lim_{x\to a}g(x)$ both exists . Otherwise it may fail. So your mistake lies in your very first step. Dec 26, 2022 at 8:36
• @Mr.GandalfSauron I would add: both exist and are not infinity of different sign Dec 26, 2022 at 8:41

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.

• Yes it makes sense to me now, after @Mr.GandalfSauron pointed out that the limit doesn't exist, it almost clicked for me and seeing your explanation, showing that it is of the indeterminate form $\infty - \infty$ it fully makes sense. Thank you very much Dec 26, 2022 at 10:23

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

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

• Assuming the limit exists Dec 26, 2022 at 10:13
• The question was not about an alternative to Taylor or De L'Hopital Dec 26, 2022 at 10:17
• This is a fascinating solution, thank you for taking the time to give this solution, though my question was asking the fault in my solution, I appreciate your answer, it is very insightful. Thank you Dec 26, 2022 at 10:22

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 $$\boxed{l=\lim_{x \to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}}$$

• The question was not about an alternative to Taylor or De L'Hopital Dec 26, 2022 at 10:18
• Thank you for the alternate solution, though it wasn't the question I asked, I appreciate your insight, it is very helpful. Dec 26, 2022 at 10:21
• @enzotib I know, I just suggested an alternative method Dec 26, 2022 at 17:55