$$\lim_{x \to 0} \frac{\sin x-x}{x^2}$$ 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 ?

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

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    $\begingroup$ This is pretty slick. $\endgroup$ – 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, $\sin(x)/x\to 1$.

  • $\begingroup$ 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? $\endgroup$ – Sewer Keeper Jun 4 '20 at 19:48
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    $\begingroup$ Near $x=0$, everything is positive. So $1-\cos(x) \geq 1-\sin(x)/x$. $\endgroup$ – 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.

We start with the trigonometric identity

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

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


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


And we are done!

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

  • $\begingroup$ @cookiemaster Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – 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 $0$.

  • $\begingroup$ Is there any algabraic way for high school level? $\endgroup$ – Cookiemaster Jun 4 '20 at 19:30
  • $\begingroup$ I don't think so. $\endgroup$ – José Carlos Santos Jun 4 '20 at 20:10
  • $\begingroup$ @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. ;-) $\endgroup$ – Mark Viola Jun 7 '20 at 18:53
  • $\begingroup$ @MarkViola (+1) Very nice! $\endgroup$ – José Carlos Santos Jun 7 '20 at 19:24
  • $\begingroup$ Thank you my friend! Much appreciated. $\endgroup$ – Mark Viola Jun 7 '20 at 20:32

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