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It is easy to see that $$\lim_{x\to 0} \frac{\sin x - x}{x^2} =0, $$but I can't figure out for the life of me how to argue without using L'Hospital or Taylor. Any ideas?

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    $\begingroup$ Note that $(\sin x-x)/x^2$ is an odd function so if a limit exists at zero then it must be its own negative, hence zero. But in order to actually show that a limit does exist, one can ask - what are you taking as the definition of the sine function if you're not allowing Taylor expansion? $\endgroup$ – Shalop Mar 25 '17 at 1:18
  • $\begingroup$ See math.stackexchange.com/questions/387333/… $\endgroup$ – lab bhattacharjee Mar 25 '17 at 2:07
  • $\begingroup$ Here's how you can do the more difficult case with $x^3$ in the denominator: math.stackexchange.com/a/158134/1242 $\endgroup$ – Hans Lundmark Mar 25 '17 at 15:36
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In THIS ANSWER, I used the integral definition of the arcsine function to show that for $0 \le x\le \pi/2$, we have the inequalities

$$x\cos(x)\le \sin(x)\le x \tag 1$$

Using the trigonometric identity $1-\cos(x)=2\sin^2(x/2)$, we see from $(1)$ that

$$-2x\,\,\underbrace{\left(\frac{\sin^2(x/2)}{x^2}\right)}_{\to \frac14}\le \frac{\sin(x)-x}{x^2}\le 0 \tag2$$

Applying the squeeze theorem to $(2)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\sin(x)-x}{x^2}=0}$$

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  • $\begingroup$ There are geometric ways to prove these inequalities, too. The easier is that $\sin t\leq t$, since $\sin t$ is the length of the shortest path from $(\cos t,\sin t)$ to the $x$-axis, and $t$ is the path along the arc of the circle from the same point to the $x$-axis. The other inequality is a little harder to show, but is essentially a geometric result. $\endgroup$ – Thomas Andrews Mar 25 '17 at 2:03
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    $\begingroup$ @ThomasAndrews Indeed. Always nice to see you're still here! There are a variety of ways to show the inequalities in $(1)$. I've referenced an answer that I posted, in which I used the integral definition of the arcsine to arrive at $(1)$. And on that page there are 42 other answers to the posted question. -Mark $\endgroup$ – Mark Viola Mar 25 '17 at 3:20
  • $\begingroup$ Applying the squeeze theorem to (2) instead of (3) @Dr.MV $\endgroup$ – Rafael Wagner Mar 25 '17 at 22:01
  • $\begingroup$ @RafaelWagner Thank you! I've edited. $\endgroup$ – Mark Viola Mar 25 '17 at 22:01
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Use the answer here and

$$\lim_{x\to 0} \frac{\sin x - x}{x^2} = -\lim_{x \to 0}x\left(\frac{x - \sin(x)}{x^3}\right) = -\lim_{x \to 0}x\lim_{x \to 0}\left(\frac{x - \sin(x)}{x^3}\right) = \frac{-1}{6}\cdot 0 = 0$$

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If you know the function $h(x)=\frac{\sin{x}}{x}$ is analytic, then $$h^{\prime}(0)=\lim_{x\to0}\frac{h(x)-h(0)}{x}=0$$ since it is an even function.

Then we obtain the desired limit.

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For $x>0,$

$$\tag 1 0\le x-\sin x = \int_0^x(1-\cos t)\, dt.$$

Now $1-\cos t \le t^2$ for all $t.$ Why? Because both sides are even, the inequality is true for $t=0,$ and it's true for the derivatives on $[0,\infty).$ Thus the right side of $(1)$ is bounded above by

$$\int_0^xt^2\, dt = x^3/3.$$ That is enough to show $\lim_{x\to 0^+}(x-\sin x)/x^2$ is $0,$ and since this function is odd, $\lim_{x\to 0^-}(x-\sin x)/x^2$ is also $0.$

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