$\lim\limits_{x \to 0} \frac{\sin x-x}{x^2}$ without L'Hospital's rule

$$\lim_{x \to 0} \frac{\sin x-x}{x^2}$$ I know it's an easy limit since inside the limit approaches $$0/0$$ as $$x$$ approaches $$0$$, we can use L'Hospital's twice to get $$\lim_{x \to 0} -\frac{\sin x}{2} = 0$$ So my question is that: Is there any way to calculate the limit other than L'Hospital's rule ?

• Yes: Taylor-Maclaurin's formula at order $3$ and some elementary asymptotic analysis. – Bernard Jun 4 '20 at 19:21
• Edit : i need a high school level way if possible – Cookiemaster Jun 4 '20 at 19:32
• It's the derivative of the sinc function $\sin(x)/x$ at $x=0$. – Mark Viola Jun 7 '20 at 18:29

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 : \begin{aligned} \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} \end{aligned}

Which means $$\left(\forall x\in\mathbb{R}^{*}\right),\ \left|\frac{x-\sin{x}}{x^{2}}\right|\leq\frac{\left|x\right|}{6}$$, and thus $$\lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 \cdot$$

• This is pretty slick. – FearfulSymmetry Jun 4 '20 at 20:31

$$\lim_{x\to 0} \left|\frac{\sin(x)-x}{x^2} \right|= \lim_{x\to 0} \left|\frac{\frac{\sin(x)}{x}-1}{x} \right|$$By some elementary geometric inequalities, for $$x$$ near $$0$$ we have $$\cos(x)\leq \frac{\sin(x)}{x}\leq 1$$Thus $$\lim_{x\to 0} \frac{1-1}{x}\leq \lim_{x\to 0}\left| \frac{\frac{\sin(x)}{x}-1}{x}\right| \leq \lim_{x\to 0} \left|\frac{{1-\cos(x)}}{x}\right|$$ $$0\leq \lim_{x\to 0}\left| \frac{\frac{\sin(x)}{x}-1}{x}\right| \leq \lim_{x\to 0} \left|\frac{\sin^2(x)}{x(1+\cos(x))}\right|$$This last limit goes to $$0$$, since by the elementary inequality above, $$\sin(x)/x\to 1$$.

• Why does the implication $$\cos x \le \frac{ \sin x }{x} \le 1 \implies \left| \frac{\sin x}{x} - 1 \right| \le\left| 1 - \cos x \right|$$ hold? – Sewer Keeper Jun 4 '20 at 19:48
• Near $x=0$, everything is positive. So $1-\cos(x) \geq 1-\sin(x)/x$. – FearfulSymmetry Jun 4 '20 at 19:51

I thought it might be of interest to present a way forward that avoids using calculus and relies instead on pre-calculus tools only. To that end we proceed.

$$\sin(3x)=3\sin(x)-4\sin^3(x) \tag 1$$

Next, we enforce the substitution $$x\to x/3$$ in $$(1)$$ yields

$$\sin(x)=3\sin(x/3)-4\sin^3(x/3)$$

Upon the subsequent iteration we obtain

$$\sin(x)=3^2\sin(x/3^2)-4\times 3^1\sin^3(x/3^2)-4\sin^3(x/3^1)$$

We have then after $$n$$ iterations

$$\sin(x)=3^n\sin(x/3^n)-4\sum_{k=1}^n3^{k-1}\sin^3(x/3^k) \tag 2$$

Using $$\sin (x)\le x$$ for $$x>0$$ in $$(2)$$ reveals

$$\sin(x)\ge 3^n\sin(x/3^n)-4x^3\sum_{k=1}^n 3^{k-1}/3^{3k}$$

Letting $$n\to \infty$$ yields

$$\sin(x)\ge x-\frac16x^3$$

for $$x>0$$. Hence, for $$x>0$$ we see that

$$-\frac16 x\le \frac{\sin(x)-x}{x^2}\le 0\tag3$$

Application of the squeeze theorem to $$(3)$$ reveals

$$\lim_{x\to 0^+}\frac{\sin(x)-x}{x^2}=0$$

Now, use the analogous development to show that the limit from the left is $$0$$ to yield the coveted result

$$\lim_{x\to0}\frac{\sin(x)-x}{x^2}=0$$

And we are done!

Tools Used: Trigonometric Series, Summation of a Geometric Series, The Squeeze Theorem

• @cookiemaster Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Aug 18 '20 at 2:25

Yes: use the fact that$$(\forall x\in\Bbb R):\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$to deduce that$$(\forall x\in\Bbb R\setminus\{0\}):\frac{\sin(x)-x}{x^2}=-\frac x{3!}+\frac{x^3}{5!}-\cdots$$So, your limit is $$0$$.

• Is there any algabraic way for high school level? – Cookiemaster Jun 4 '20 at 19:30
• I don't think so. – José Carlos Santos Jun 4 '20 at 20:10
• @JoséCarlosSantos Hi my friend. I hope you are staying safe and healthy. There is a way forward that circumvents use of differential calculus. I've posted an approach on this page. Let me know if you like it. ;-) – Mark Viola Jun 7 '20 at 18:53
• @MarkViola (+1) Very nice! – José Carlos Santos Jun 7 '20 at 19:24
• Thank you my friend! Much appreciated. – Mark Viola Jun 7 '20 at 20:32