Regarding $\lim \limits_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}$ 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?
What you did is wrong, remember that
$$|\sin x|=\begin{cases}\sin x&amp;if\ x\in[0,\pi]\\ -\sin x&amp;if\ x\in[0,-\pi]\end{cases}.$$

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.
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?