Mathematics - Continuity - Differentiability Question with Solution | TestHub

For $f(x)$ to be continuous at $x=0$, we must have ${\lim }_{x\to 0}f(x)=f(0)$. Given $f(0)=\frac{1}{2}$. We evaluate the limit: ${\lim }_{x\to 0}\frac{1-\cos (kx)}{x\sin x}={\lim }_{x\to 0}\frac{2{\sin }^2(\frac{kx}{2})}{x\sin x}$. Multiplying and dividing by appropriate terms, we get ${\lim }_{x\to 0}\frac{2{\sin }^2(\frac{kx}{2})}{(\frac{kx}{2}{)}^2}\cdot \frac{(\frac{kx}{2}{)}^2}{x\frac{\sin x}{x}\cdot x}={\lim }_{x\to 0}2\cdot 1\cdot \frac{k^2{x}^2/4}{x^2\cdot 1}=\frac{2k^2}{4}=\frac{k^2}{2}$. Equating this to $f(0)$, we have $\frac{k^2}{2}=\frac{1}{2}$, which implies $k^2=1$. Thus, the value of $k^2$ is 1.