Question

If $f\left(x\right)=\{\begin{array}{c}\frac{1-\cos x}{x};x\ne 0\\ k;x=0\end{array}$ 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 that, $f\left(x\right)=\{\begin{array}{c}\frac{1-\cos x}{x};x\ne 0\\ k;x=0\end{array}$

Now, It is given that $f\left(x\right)$ is continuous at $x=0$

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

Using L'Hospital's rule, we get

$\underset{x\to 0}{lim}\sin x=k$

$\Rightarrow 0=k$

Hence, the value of $k$ is $0$.