Given, $\lim _\limits{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _\limits{x \rightarrow 0} \frac{1-1+2 \sin ^2 \frac{m x}{2}}{1-1+2 \sin ^2 \frac{n x}{2}}$ $\left[\begin{array}{l}\because \quad \cos m x=1-2 \sin ^2 \frac{m x}{2} \\ \text { and } \sin n x=1-2 \sin ^2 \frac{n x}{2}\end{array}\right]$

$$=\lim _\limits{x \rightarrow 0} \frac{\sin ^2 \frac{m x}{2}}{\sin ^2 \frac{n x}{2}}=\lim _\limits{x \rightarrow 0} \frac{\frac{\sin ^2 \frac{m x}{2}}{\left(\frac{m x}{2}\right)^2} \cdot\left(\frac{m x}{2}\right)^2}{\frac{\sin ^2 \frac{n x}{2}}{\left(\frac{n x}{2}\right)^2} \cdot\left(\frac{n x}{2}\right)^2}=\frac{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right)^2}{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right)^2} \cdot \frac{m^2 \frac{x^2}{4}}{n^2 \frac{x^2}{4}}$$

$$=\frac{m^2}{n^2} \cdot \frac{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right)^2}{\lim _\limits{x \rightarrow 0}\left(\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right)^2}=\frac{m^2}{n^2} \quad\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]$$

$[\because x\to0\Rightarrow k ~x\to 0]$

Given, $\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{1-1+2 \sin ^2 3 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}\quad \left[\because \cos 2 x=1-2 \sin ^2 x\right]$

$$\begin{array}{ll} =\lim _\limits{x \rightarrow \pi / 3} \frac{\sqrt{2} \sin 3 x}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _\limits{x \rightarrow \pi / 3} \frac{\sin 3 x}{\frac{\pi}{3}-x} & \\ =\lim _\limits{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{\frac{\pi-3 x}{3}} & {[\because \sin (\pi-\theta)=\sin \theta]} \\ =3 \lim _\limits{x \rightarrow \pi / 3} \frac{\sin (\pi-3 x)}{(\pi-3 x)}=3 \times 1 & {\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]} \\ =3 & {\left[\because x \rightarrow \frac{\pi}{3} \Rightarrow\left(x-\frac{\pi}{3}\right) \rightarrow 0\right]} \end{array}$$

Given, $\lim _\limits{x \rightarrow \pi / 4} \frac{\sqrt{2}\left(\sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}\right)}{\left(x-\frac{\pi}{4}\right)}=\lim _\limits{x \rightarrow \pi / 4} \frac{\sqrt{2}\left(\sin x \cos \frac{\pi}{4}-\cos x \cdot \sin \frac{\pi}{4}\right)}{\left(x-\frac{\pi}{4}\right)}$

$$\begin{array}{ll} =\lim _\limits{x \rightarrow \pi / 4} \frac{\sqrt{2}\left\{\sin \left(x-\frac{\pi}{4}\right)\right\}}{\left(x-\frac{\pi}{4}\right)} \\ =\sqrt{2} \lim _\limits{x \rightarrow \pi / 4} \frac{\sin \left(x-\frac{\pi}{4}\right)}{\left(x-\frac{\pi}{4}\right)}=\sqrt{2} & {\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]} \\ & {\left[\because x \rightarrow \frac{\pi}{4} \Rightarrow\left(x-\frac{\pi}{4}\right) \rightarrow 0\right]} \end{array}$$

Given, $\lim _\limits{x \rightarrow \pi / 6} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}=\lim _\limits{x \rightarrow \pi / 6} \frac{2\left(\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right)}{\left(x-\frac{\pi}{6}\right)}$

$$\begin{array}{r} =\lim _\limits{x \rightarrow \pi / 6} \frac{2\left(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)}=2 \lim _\limits{x \rightarrow \pi / 6} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)} \\ =2 \quad[\because \sin A \cos B-\cos A \sin B=\sin (A-B)] \\ {\left[\because \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1\right]} \\ {\left[\because x \rightarrow \frac{\pi}{6} \Rightarrow\left(x-\frac{\pi}{6}\right) \rightarrow 0\right]} \end{array}$$

$$\begin{aligned} &\text { Given, }\\ &\begin{aligned} \lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x} & =\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x+3 x}{2 x} \cdot 2 x}{\frac{2 x+\tan 3 x}{3 x} \cdot 3 x} \\ & =\lim _{x \rightarrow 0} \frac{\left(\frac{\sin 2 x}{2 x}+\frac{3 x}{2 x}\right) 2 x}{\left(\frac{2 x}{3 x}+\frac{\tan 3 x}{3 x}\right) 3 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{2 x}+\frac{3}{2}}{\frac{2}{3}+\frac{\tan 3 x}{3 x}} \cdot \frac{2}{3} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{2}{3} \lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{2 x}+\frac{3}{2}}{\frac{2}{3}+\lim _{x \rightarrow 0} \frac{\tan 3 x}{3 x}} \\ & =\frac{2}{3}\left(\frac{1+\frac{3}{2}}{\frac{2}{3}+1}\right) \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right] \\ & =\frac{2}{3} \times \frac{\frac{5}{2}}{\frac{5}{3}}=\frac{2}{3} \times \frac{5}{2} \times \frac{3}{5}=1 \end{aligned}$$