Question

Find the value of $k$ if $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos 2x}{1+\cos 2x},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0$.

Answer 1

$\Rightarrow \underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=k$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{1-\cos 2x}{1+\cos 2x}$

$=\underset{x\to 0}{lim}\frac{1-\cos 2(0-h)}{1+\cos 2(0-h)}$

$=\underset{x\to 0}{lim}\frac{1-\cos 2h}{1+\cos 2h}\left(\cos \right(-x)=\cos x)$

$=\frac{1-1}{1+1}$

$=0$

$\underset{x\to 0}{lim}f\left(x\right)=0=k$

$\therefore k=0$