Question

Show that the function is continuous at x =0, if
${lim}_{x\to 0}=\frac{\sin 3x}{\tan 2x}$

Answer 1

${lim}_{x\to 0^{-}}=\frac{\sin 3x}{\tan 2x}\Rightarrow \left(\frac{\sin 3x}{3x}\right)\times \left(\frac{2x}{\tan 2x}\right)\times \frac{3}{2}\Rightarrow 1\times 1\times \frac{3}{2}\Rightarrow \frac{3}{2}{lim}_{x\to 0}\frac{\sin 3x}{\tan 2x}\Rightarrow \frac{\sin 3x}{3x}\times \frac{2x}{\tan 2x}\times \frac{3}{2}\Rightarrow \frac{3}{2}Hence{lim}_{x\to 0^{-}}={lim}_{x\to 0}iscontinuous.$