Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x\log \cos x}{\log (1+x^2)},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$ then

• A. $f\left(x\right)$ is not continuous at $x=0$
• B. $f\left(x\right)$ is continous at $x=0$
• C. $f\left(x\right)$ is continous at $x=0$ but not differentiable at $x=0$
• D. None of these

Answer 1

The correct option is B $f\left(x\right)$ is continous at $x=0$

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

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

$=\underset{x\to 0}{lim}\frac{\log [1-(1-\cos x)]}{1-\cos x}.\frac{1-\cos x}{\log (1+x^2)}$

$=-\underset{x\to 0}{lim}\frac{[1-(1-\cos x)]}{-(1-\cos x)}.\frac{2{\sin }^2\frac{x}{2}}{4{\left(\frac{x}{2}\right)}^2}.\frac{x^2}{\log (1+x^2)}=-\frac{1}{2}$

Hence $f\left(x\right)$ is differentiable at $x=0$ and so continuous at $x=0$