Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $0$

Given, $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos x}{x},& x\ne 0\\ k,& x=0\end{array}$
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{1-\cos x}{x}$
$=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}}{4{\left(\frac{x}{2}\right)}^2}\times x=0$
Also, $f\left(0\right)=k$
$\because f\left(x\right)$ is continuous
$\therefore \underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
$\Rightarrow k=0$.