**Question:**

> 
> Find the limit of: $$\lim\_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}$$
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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?