Question

$\underset{x\to 0}{lim}\frac{\sqrt{1-\cos x}}{x}$ is equal to

• A. $-\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $0$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

We know $\sqrt{1-\cos x}=\{\begin{array}{cc}-\sqrt{2}\sin \frac{x}{2},& x<0\\ \sqrt{2}\sin \frac{x}{2},& x\ge 0\end{array}$
Therefore, $\underset{x\to 0^{-}}{lim}\frac{\sqrt{1-\cos x}}{x}=\underset{x\to 0^{-}}{lim}-\frac{\sqrt{2}\sin \frac{x}{2}}{x}=-\frac{1}{\sqrt{2}}$
and $\underset{x\to 0^{+}}{lim}\frac{\sqrt{1-\cos x}}{x}=\underset{x\to 0^{+}}{lim}\frac{\sqrt{2}\sin \frac{x}{2}}{x}=\frac{1}{\sqrt{2}}$
Since, $\underset{x\to 0^{-}}{lim}\frac{\sqrt{1-\cos x}}{x}\ne \underset{x\to 0^{+}}{lim}\frac{\sqrt{1-\cos x}}{x}$
So, $\underset{x\to 0}{lim}\frac{\sqrt{1-\cos x}}{x}$ does not exist.