## Thomas' Calculus 13th Edition

$$\lim _{x \rightarrow 0}\frac{x(1-\cos x)}{\sin ^2 3 x} =0$$
Given $$\lim _{x \rightarrow 0}\frac{x(1-\cos x)}{\sin ^2 3 x}$$ So, we get \begin{aligned} L&=\lim _{x \rightarrow 0}\frac{x(1-\cos x)}{\sin ^2 3 x} \\ &= \lim _{x \rightarrow 0}\frac{x(1-\cos x)}{\sin 3 x \cdot \sin 3 x}\\ &= \lim _{x \rightarrow 0}\left(\frac{ x}{\sin 3 x}\right)\left(1-\cos x\right)\left(\frac{1}{\sin 3 x}\right)\\ &= \lim _{x \rightarrow 0}\left(\frac{1}{3}\right)\left(\frac{3 x}{\sin 3 x}\right)\left(\frac{1-\cos x}{x}\right)\left(\frac{ x}{\sin 3 x}\right)\\ &= \left(\frac{1}{3}\right)\lim _{x \rightarrow 0}\left(\frac{3 x}{\sin 3 x}\right)\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x}\right)\lim _{x \rightarrow 0}\left(\frac{ x}{\sin 3 x}\right)\\ &= \frac{1}{3}\lim _{x \rightarrow 0}\left(\frac{3 x}{\sin 3 x}\right)\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x}\right)\lim _{x \rightarrow 0} \frac{1}{3}\left(\frac{3 x}{\sin 3 x}\right)\\ &=\frac{1}{3}\cdot 0\cdot\frac{1}{3} \\ &=0 \end{aligned} Where, \begin{aligned} \lim _{x \rightarrow 0} \frac{1-\cos x}{x}&= \lim _{x \rightarrow 0} \frac{1-\cos x}{x}\frac{1+\cos x}{1+\cos x}\\ &= \lim _{x \rightarrow 0} \frac{1-\cos^2 x}{x(1+\cos x)}\\ &= \lim _{x \rightarrow 0} \frac{\sin^2 x}{x(1+\cos x)}\\ &= \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot\lim _{x \rightarrow 0} \sin x \cdot\lim _{x \rightarrow 0} \frac{1}{(1+\cos x)}\\ &= 1\cdot\sin 0\cdot \frac{1}{(1+\cos 0)}\\ &=0 \end{aligned}