Question

Evaluate the following limits:

$\underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sqrt{1-\cos 6x}}{\sqrt{2}\left({\displaystyle \frac{\mathrm{\pi }}{3}}-x\right)}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sqrt{1-\cos 6x}}{\sqrt{2}\left({\displaystyle \frac{\mathrm{\pi }}{3}}-x\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sqrt{2{\sin }^{2}3x}}{\sqrt{2}\left({\displaystyle \frac{\mathrm{\pi }}{3}-x}\right)}\left(1-\cos 2\theta =2{\sin }^2\theta \right) \\ =\underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sqrt{2}\sin 3x}{\sqrt{2}\left({\displaystyle \frac{\mathrm{\pi }}{3}-x}\right)} \\ =\underset{x\to \frac{\mathrm{\pi }}{3}}{\lim }\frac{\sin 3x}{\left({\displaystyle \frac{\mathrm{\pi }}{3}-x}\right)} \end{gathered}$$
$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{\sin 3\left({\displaystyle \frac{\mathrm{\pi }}{3}}+h\right)}{{\displaystyle \frac{\mathrm{\pi }}{3}}-\left({\displaystyle \frac{\mathrm{\pi }}{3}}+h\right)}\left(\mathrm{Put}x=\frac{\mathrm{\pi }}{3}+h\right) \\ =\underset{h\to 0}{\lim }\frac{\sin \left(\mathrm{\pi }+3h\right)}{{\displaystyle -h}} \\ =\underset{h\to 0}{\lim }\frac{-\sin 3h}{{\displaystyle -h}}\left[\sin \left(\mathrm{\pi }+\theta \right)=-\sin \theta \right] \\ =3\times \underset{h\to 0}{\lim }\frac{\sin 3h}{{\displaystyle 3h}} \\ =3\times 1\left(\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1\right) \\ =3 \end{gathered}$$