Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin 2x-\tan 2x}{(e^x-1)x^2},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0,$ then the absolute value of $k$ is

Answer 1

For $f\left(x\right)$ to be continuous at $x=0$
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
Hence, $k=\underset{x\to 0}{lim}\frac{\sin 2x-\tan 2x}{(e^x-1)\cdot x^2}$
$\Rightarrow k=\underset{x\to 0}{lim}\frac{\sin 2x-\frac{\sin 2x}{\cos 2x}}{(e^x-1)\cdot x^2}\Rightarrow k=\underset{x\to 0}{lim}\frac{\sin 2x(\cos 2x-1)}{\cos 2x(e^x-1)\cdot x^2}\Rightarrow k=-\underset{x\to 0}{lim}[\left(\frac{\sin 2x}{2x}\right)\left(\frac{2{\sin }^{2}x}{x^2}\right)\left(\frac{1}{\cos 2x}\right)\left(\frac{x}{e^x-1}\right)\cdot 2]\Rightarrow k=-1\cdot 2\cdot 1\cdot 1\cdot \left(2\right)\Rightarrow k=-4\Rightarrow |k|=4$