Question

$\underset{x\to 0}{\lim }\frac{e^x-1}{\sqrt{1-\cos x}}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{e^x-1}{\sqrt{1-\cos x}}\right] \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator},\mathrm{we}\mathrm{get}: \\ =\underset{x\to 0}{\lim }\left[\frac{\left(e^x-1\right)}{\sqrt{1-\cos x}}\times \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(e^x-1\right)\left(\sqrt{1+\cos x}\right)}{\sqrt{1-{\cos }^{2}x}}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(e^x-1\right)\sqrt{1+\cos x}}{\left|\sin x\right|}\right] \end{gathered}$$

Dividing numerator and the denominator by x, we get:

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\left(\frac{e^x-1}{x}\right)\times \frac{\sqrt{1+\cos x}}{\left({\displaystyle \frac{\left|\sin x\right|}{x}}\right)}\right] \\ \mathrm{Left}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{-}}{\lim }\left[\left(\frac{e^x-1}{x}\right)\times \frac{\sqrt{1+\cos x}}{\left({\displaystyle \frac{\left|\sin x\right|}{x}}\right)}\right] \\ =\underset{x\to 0^{-}}{\lim }\left[\left(\frac{e^x-1}{x}\right)\times \frac{\sqrt{1+\cos x}}{\left({\displaystyle \frac{-\sin x}{x}}\right)}\right] \\ =-\frac{1\times \sqrt{2}}{1} \\ =-\sqrt{2} \\ \mathrm{Right}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{+}}{\lim }\left[\left(\frac{e^x-1}{x}\right)\times \frac{\sqrt{1+\cos x}}{{\displaystyle \frac{\left|\sin x\right|}{x}}}\right] \\ =\underset{x\to 0^{+}}{\lim }\left[\left(\frac{e^x-1}{x}\right)\times \frac{\sqrt{1+\cos x}}{{\displaystyle \frac{\sin x}{x}}}\right] \\ =1\times \frac{\sqrt{2}}{1} \\ =\sqrt{2} \end{gathered}$$
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Left hand limit ≠ Right hand limit
Thus, limit does not exist.