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Calculate the limit without using de l'Hopital:
$$\lim_{x\to 0} \frac{x-\sin x} {1-\cos x}$$ I want to use the limit:$$\lim_{x\to 0} \frac{\sin x}{x}=1$$ but I don't know how to do it.

I manipulated the expression to get $$\lim_{x\to 0} \frac{x}{x-\sin x}-\lim_{x\to 0} \frac{\sin x} {1-\cos x}$$ but I don't know where to go from here.

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  • $\begingroup$ Can you use Taylor series? $\endgroup$ – KittyL Mar 29 '15 at 10:06
  • $\begingroup$ You can't simply use $\lim_{x\to0}\frac{\sin x}{x}=1$, because the limit depends on the cubic term of the Taylor development of $\sin x$. So either you know about $x-\sin x\le x^3/6$ or you need Taylor developments. And using them is much stronger than using l'Hôpital's theorem. $\endgroup$ – egreg Mar 29 '15 at 10:39
  • $\begingroup$ @egreg Your claim in the comment above is not true. You can clearly prove it from the definition of $\sin$ and either: squeeze theorem, nested intervals, intermediate values theorem for continuous functions, or even continuous induction. The reason is that Lagrange's theorem is equivalent to any of these, and Taylor, and L'Hospital are equivalent to Lagrange's theorem. $\endgroup$ – Nathanson Mar 29 '15 at 10:42
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    $\begingroup$ @egreg Moreover, I think that there is already an answer in this website exactly or that implies $\frac{x-\sin(x)}{x^2}\to0$ without L'Hospital. I would be good to try to find it. This type of tour-de-force exercise is not really worth it of expending the time to recreate it. $\endgroup$ – Nathanson Mar 29 '15 at 10:48
  • $\begingroup$ @Nathanson I was saying that just the basic limit is not sufficient and some further knowledge is needed, in whatever way you get it. In my opinion, such assignments are silly: using the appropriate tool (be it l'Hôpital or Taylor, for this case) shows better if one knows the subject. $\endgroup$ – egreg Mar 29 '15 at 11:02
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$f(x)=\frac{x-\sin(x)}{1-\cos(x)}$ is an odd function, hence as soon as we prove that $f(x)$ is continuous at the origin we have $\lim_{x\to 0}f(x)=0$. On the other hand, for any $t\in\left[0,\frac{\pi}{2}\right]$ we have $$\frac{2}{\pi} t\leq \sin(t) \leq t \tag{1} $$ from the concavity of the sine function. By integrating over $(0,x)$, with $x\in\left[0,\frac{\pi}{2}\right]$, we get: $$ \frac{x^2}{\pi}\leq 1-\cos(x) \leq \frac{x^2}{2}\tag{2} $$ and by integrating again: $$ \frac{x^3}{3\pi}\leq x-\sin(x) \leq \frac{x^3}{6}\tag{3} $$ By $(2)$ and $(3)$ we have that in a right neighbourhood of the origin the ratio $\frac{x-\sin(x)}{1-\cos x}$ is bounded between $\frac{2}{3\pi}x$ and $\frac{\pi}{6}x$. It follows that $f(x)$ is (Lipschitz-)continuous at the origin and $\lim_{x\to 0}f(x)=0$.

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Lemma: If $f$ is continuous, nonnegative, and increasing on some $[0,b],$ then $0\le \int_0^x f(t)\,dt \le x\cdot f(x)$ for $x\in [0,b].$ Proof: Obvious.

Now $x - \sin x = \int_0^x (1-\cos t)\, dt.$ Since $1-\cos t$ is continuous, nonnegative, and increasing on $[0,\pi],$ we see by the lemma that $0\le x-\sin x \le x(1-\cos x)$ for $x\in [0,\pi].$ For $x\in (0,\pi]$ we then have

$$0\le \frac{x-\sin x}{1-\cos x} \le \frac{x(1-\cos x)}{1-\cos x} = x.$$

Thus our limit from the right equals $0.$ Because we're dealing with an odd function, the limit from the left is $0$ also, showing that the limit of interest is $0.$

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First of all you can note that as $x \to 0^{+}$ we have $$\sin x < x < \tan x$$ and therefore $$0 < x - \sin x < \tan x - \sin x = \tan x (1 - \cos x)$$ or $$0 < \frac{x - \sin x}{1 - \cos x} < \tan x$$ and applying Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{x - \sin x}{1 - \cos x} = 0$$ The case $x \to 0^{-}$ is now easily handled by putting $x = -t$ and noting that $t \to 0^{+}$.

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