# Finding $\lim_{x\rightarrow 0} \frac{\sin(x)-x}{x^2}$

I'm trying to find the following limit:

$$\lim_{x\rightarrow 0} \frac{\sin(x)-x}{x^2}$$

I'm not allowed to use L'Hospital's rule this moment. I tried to use the tools I have, which are $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$, squeeze and make substitutions, but without success. I'm asking here only a hint if it's possible. Thanks!

This is a very nice limit. Let me derive the crucial limit here.

$$L=\lim_{x\to0}\frac{\sin x-x}{x^3}$$

Now we will substitute $$x=2x$$ $$L=\lim_{x\to0}\frac{\sin 2x-2x}{8x^3}$$ $$4L=\lim_{x\to0}\frac{\frac12\sin 2x-x}{x^3}$$ We will substract from this our original limit $$4L-L=3L=\lim_{x\to0}\frac{\frac12\sin 2x-\sin x}{x^3}$$ $$3L=\lim_{x\to0}\frac{\sin{x}\cos{x}-\sin x}{x^3}$$ $$3L=\lim_{x\to0}\frac{\sin{x}(\cos{x}-1)}{x^3}$$ $$3L=\lim_{x\to0}\frac{\sin{x}(\cos{x}-1)(\cos{x}+1)}{x^3(\cos{x}+1)}$$ $$3L=\lim_{x\to0}\frac{-\sin^3{x}}{x^3(\cos{x}+1)}$$ $$3L=\lim_{x\to0}\frac{-1}{(\cos{x}+1)}$$ $$3L=-\frac{1}{2}$$ $$L=-\frac{1}{6}$$

And now a hint for your specific limit:

$$\lim_{x\to0}\frac{\sin{x}-x}{x^2}=\lim_{x\to0}x\frac{\sin{x}-x}{x^3}$$

• Very nice!! Thanks! This was great and used simple algebra. But of course, this looks like a problem that it's hard to get the answer by yourself. – Dunck Mar 25 at 15:35
• It sure must seem non-intuitive at first, but I think you get used to it, since if you solve lots of limits (without advanced techniques) you will probably use this trick many times. The reason we are doing the substitution x=2x and then substracting is to get rid of that x in the numerator. If you are interested you should definitely look at this answer: math.stackexchange.com/a/936642/726980 which has solutions to six similar limits. – Matthew Mar 25 at 15:48
• I now see that it was unneccessary to solve my (not-so-crucial) limit first and that you could easily do your limit directly in the similar way. You can try that as an exercise, – Matthew Mar 25 at 15:57
• @Dunck Note that this solution assumes the existence of the limit $L$ without proving it. You can check this answer for a similar derivation except the existence assumption is not there. It gets a lot more complicated then. – bjorn93 Mar 25 at 16:41
• @Dunck In general, you can't simply assume the existence of a limit without proving it. In this case, the limit $L$ exists, and you get a simple solution. But you could "prove" a wrong limit using this strategy – bjorn93 Mar 25 at 17:01

We can show that $$|\sin(x)-x|\leq \frac{|x|^3}{6}$$ for all $$x\in\mathbb{R}$$. Consider the function $$f(x)=\frac{|x|^3}{6}-|\sin(x)-x|$$. We have to show that $$f(x)\geq 0$$. Notice that $$f$$ is even, i.e. $$f(-x)=f(x)$$, so it suffices to prove this for $$x\geq 0$$. Then $$x\geq \sin(x)$$, and $$f(x)=\frac{x^3}{6}-(x-\sin(x))=\frac{x^3}{6}-x+\sin(x)$$ for $$x\geq 0$$. The derivative is $$f'(x)=\frac{x^2}{2}-1+\cos(x)\geq 0$$, so $$f(x)\geq f(0)=0$$ for $$x\geq 0$$ which proves the inequality. Thus, $$\frac{|\sin(x)-x|}{x^2}\leq \frac{|x|}{6} \Leftrightarrow \\ -\frac{|x|}{6}\leq \frac{\sin(x)-x}{x^2} \leq \frac{|x|}{6}$$ You can conclude with the squeeze theorem now.

• I think that as soon as you took the derivative, you implicitly used L'hospital. – B. Goddard Mar 24 at 21:06
• @B.Goddard No, the derivatives of the elementary functions can be proved using basic tools. L'Hospital's rule is not needed. – bjorn93 Mar 24 at 21:07
• That's not what I mean. L'hospital works by approximating the functions by their Taylor polynomials. So using Taylor polynomials is essentially using L'hospital. Likewise, by taking a derivative, you're approximating the function with it's first order Taylor polynomial. – B. Goddard Mar 24 at 21:10
• @B.Goddard OK, I'm assuming derivatives are available. I believe the inequality can be shown without any calculus if that's you mean. But I think it gets too convoluted. – bjorn93 Mar 24 at 21:13
• Derivatives are avalilable, but L'Hospital not. – Dunck Mar 24 at 21:16

Let $$x\in\mathbb{R}^{*}$$, observe that : $$\fbox{\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}}$$

Using the fact that $$\left(\forall t\in\left[0,1\right]\right),\ \left|\cos{\left(tx\right)}\right|\leq 1$$, we have : $$\left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\right|\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left|\cos{\left(tx\right)}\right|\mathrm{d}t}\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t}$$

Which means $$\left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{3}}\right|\leq\frac{\left|x\right|}{6}$$, the limit would then be $$0 \cdot$$

• While clever, this answer is not likely to shed any light on such problems (or even this specific problem) for the OP. – Greg Martin Mar 24 at 20:35
• @Greg Martin I wanted to give a full answer, but he said he wanted only a hint. – CHAMSI Mar 24 at 20:37
• Actually, at this momment, I'm not supposed to know the meaning int the right thing in the equation. – Dunck Mar 24 at 20:39
• @Greg Martin see the update. – CHAMSI Mar 24 at 20:48
• @Dunck if you had never seen integrals yet. The only way left is to proof the following inequality through studying functions : $$\left(\forall x\in\mathbb{R}\right),\ x-\frac{x^{3}}{6}\leq\sin{x}\leq x-\frac{x^{3}}{6}+\frac{x^{5}}{120}$$ – CHAMSI Mar 24 at 20:53

Note that:

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$

Therefore, the given limit can be rewritten as:

$$\lim_{x\rightarrow0}\frac{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots)-x}{x^2}$$ $$=\lim_{x \rightarrow0}\frac{-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots}{x^2}$$ $$=\lim_{x \rightarrow0} (-\frac{x^1}{3!}+\frac{x^3}{5!}-\frac{x^5}{7!}+\dots)$$

Substitue $$x=0$$ in the last expression, we get:

$$-\frac{0^1}{3!}+\frac{0^3}{5!}-\frac{0^5}{7!}+\dots=0$$

We have just expanded $$\sin(x)$$ using Taylor series. You mentioned that you do not know it, but it is a useful information that $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots.$$ Keep it in mind. Hope my answer helps you.

• @Axion004 Well, you are right, I will edit my answer now. Thank you. – Hussain-Alqatari Mar 25 at 15:22

I believe this is already available on MSE but my search capabilities here and on approach0 are limited.

First deal with $$x\to 0^{+}$$. We have the famous inequality $$\sin x for $$x\in(0,\pi/2)$$. And this means that we have $$0 and dividing by $$x^2$$ we get $$0<\frac{x-\sin x} {x^2}<\frac{\tan x - \sin x} {x^2}$$ The last fraction above tends to $$0$$ and by Squeeze the desired limit is $$0$$. For $$x\to 0^{-}$$ put $$t=-x$$ and proceed.

Also note that the task of evaluating a limit includes the task of proving that the limit exists or not and finding the limit if it exists. If one evaluates limit in step by step manner using limit laws and gets a concrete answer then the steps involved prove the existence of limit involved. That how the limit laws are designed.

This is an answer given by a professor from USP (Universidade de São Paulo - Brasil). I find much more clever than the answers posted here. Let's see:

We now from circle above this happens: (first we'll make the limit $$x\rightarrow 0^+$$, next making $$x\rightarrow 0^-$$)

$$\cos x-1\leq \sin x \leq \frac{1}{\cos x} \implies \frac{\cos x-1}{x} \leq \frac{\sin x - x}{x^2} \leq \frac{1-\cos x}{x\cos x}$$

Using the squeeze theorem with $$x\rightarrow 0^+$$, it's easy to find the both limits, the one in the left and the one in the right are equals zero. After that, with $$x\rightarrow 0^+$$, we find that both limits are equals to zero as well. Then, the limit required is zero.