Question

Find the value of $k$, if
$\begin{array}{cccc}f\left(x\right)& =& \frac{1-cos2x}{1-cosx},& x\ne 0\\ & =& k,& x=0\end{array}$
is constinuous at $x=0$.

Answer 1

Given

$\begin{array}{cccc}f\left(x\right)& =& \frac{1-cos2x}{1-cosx},& x\ne 0\\ & =& k,& x=0\end{array}$ is continuous at $x=0$

$⟹{lim}_{x\to 0}f\left(x\right)=f\left(0\right)$

$⟹{lim}_{x\to 0}\frac{1-\cos 2x}{1-\cos x}=k$

$⟹{lim}_{x\to 0}\frac{1-(2{\cos }^{2}x-1)}{1-\cos x}=k$

$⟹{lim}_{x\to 0}\frac{2-2{\cos }^{2}x}{1-\cos x}=k$

$⟹{lim}_{x\to 0}\frac{2(1-{\cos }^{2}x)}{1-\cos x}=k$

$⟹{lim}_{x\to 0}\frac{2(1-\cos x)(1+\cos x)}{1-\cos x}=k$

$⟹{lim}_{x\to 0}2(1+\cos x)=k$

$⟹2(1+\cos 0)=k$

$⟹k=2(1+1)=4$

Hence, $k=4$