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Question:

Find the limit of: $$\lim_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}$$

My approach: By rationalizing the denominator, we get: $$\lim_{x \to 0} \frac{(e^x-1)(\sqrt{1+\cos x})}{\sin x}$$

Dividing both the numerator and the denominator by $x$: $$\lim_{x \to 0} \frac{\frac{(e^x-1)}{x}(\sqrt{1+\cos x})}{\frac{\sin x}{x}}$$

It is known that: $$\lim_{x \to 0}\frac{(e^x-1)}{x} = 1$$ $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$

Thus, simplifying, we get the answer to be $\sqrt2$

However, it is given that the function is divergent at $x = 0$ and thus the limit is undefined. Why is this so?

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    $\begingroup$ "Function undefined at $x=0$" doesn't necessarily mean lim divergent at $x=0$, does it? $\endgroup$ – Vim Feb 23 '15 at 13:29
  • $\begingroup$ @Vim Why not though? $\endgroup$ – Gummy bears Feb 23 '15 at 13:32
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    $\begingroup$ Your mistake is that $|\sin|=\sqrt{\sin^2}\neq \sin$, the lateral limits do not coincide. $\endgroup$ – Git Gud Feb 23 '15 at 13:32
  • $\begingroup$ @Gummybears Take a look at one simplest example, let's $f(x)=\frac{x}x$ and sure it's undefined at zero but what about $\displaystyle\lim_{x\to0}f(x)$? $\endgroup$ – Vim Feb 23 '15 at 13:45
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What you did is wrong, remember that $$|\sin x|=\begin{cases}\sin x&if\ x\in[0,\pi]\\ -\sin x&if\ x\in[0,-\pi]\end{cases}.$$

So,

$$\frac{e^x-1}{\sqrt{1-\cos x}}=\frac{(e^x-1)\sqrt{1+\cos x}}{\underbrace{\sqrt{1-\cos^2x}}_{=|\sin x|}}=\frac{(e^x-1)\sqrt{1+\cos x}}{|\sin x|}$$

and thus, $$\lim_{x\to 0^+}\frac{e^x-1}{\sqrt{1-\cos x}}=\lim_{x\to 0^+}\frac{(e^x-1)\sqrt{1+\cos x}}{\sin x}=\sqrt 2$$ and $$\lim_{x\to 0^-}\frac{e^x-1}{\sqrt{1-\cos x}}=\lim_{x\to 0^-}\frac{(e^x-1)\sqrt{1+\cos x}}{-\sin x}=-\sqrt 2.$$

Therefore $$\lim_{x\to 0}\frac{e^x-1}{\sqrt{1-\cos x}}$$ is not defined.

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Your problem is that in simplifying a square root, you should use the fact that $\sqrt{x^2}=|x|,$ not $x.$ So in your simplification you should have $|\sin x|$ instead of $\sin x.$ Since $\sin$ is an odd function, the limit from the left is negative and from the right is positive.

See https://www.desmos.com/calculator/bagsuuzsjp

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Using Taylor will also give you a good explanation:
Note this $$e^{x}=1+x+o(x)$$ And this $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)$$ at the neighbourhood of $x=0$, thus $$e^x-1=x+o(x)$$ And $$1-\cos(x)=\frac{x^2}2-\frac{x^4}{4!}+o(x^4)$$ Hence $$\lim_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}=\lim_{x\to0}\frac{x+o(x)}{\sqrt{\frac12x^2(1-\frac{x^2}{12}+o(x^2))}}=\sqrt2\lim_{x\to0}\frac{x+o(x)}{|x|\sqrt{1-\frac{x^2}{12}+o(x^2)}}$$ Then note that it can be either $x>0$ or $x<0$.I think you are certainly able to take it from here.

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Using L'Hospitals rule directly:

$\frac{e^x-1}{\sqrt{1-cos x}}=\frac{(e^x-1)'}{(\sqrt{1-cos x})'}=\frac{2 e^x \sqrt{1-cos x}}{sin x}$. Now let $x \rightarrow 0$, what do you observe?

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    $\begingroup$ I don't see how this helps. $\endgroup$ – Git Gud Feb 23 '15 at 13:33
  • $\begingroup$ That I did notice. However, my question was why I can't solve using my method. $\endgroup$ – Gummy bears Feb 23 '15 at 13:34
  • $\begingroup$ You can't conclude with l'Hopital. $\endgroup$ – Surb Feb 23 '15 at 13:41

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