Question

$\underset{x\to 0}{lim}\frac{\ell n\left(\sin 3x\right)}{\ell n\left(\sin x\right)}$ is equal to

Answer 1

${lim}_{x\to 0}\frac{ln\left(sin3x\right)}{ln\left(sinx\right)}$ (using L'Hospital rule)

${lim}_{x\to 0}\frac{\frac{d}{dx}\left(lnsin3x\right)}{\frac{d}{dx}\left(lnsinx\right)}={lim}_{x\to 0}\frac{\frac{3cos3x}{sin3x}}{\frac{cosx}{sinx}}\Rightarrow {lim}_{x\to 0}\frac{3.cot3x}{cotx}$

$3{lim}_{x\to 0}\frac{cos3x.sinx}{cosx.sin3x}=3{lim}_{x\to 0}\frac{(4co{s}^{3}x-3cosx).sinx}{cosx.(3sinx-4si{n}^{3}x)}$

$\Rightarrow 3{lim}_{x\to 0}\frac{cosx(4co{s}^{2}x-3).sinx}{cosx.sinx(3-4si{n}^{2}x)}$

using limit

$\Rightarrow 3\left[\frac{4\times 1-3}{3-4\times 0}\right]=\frac{3\times 1}{3}=1$