Question

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

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is continuous at $x=0$ but not differentiable at $x=0$
• C. $f\left(x\right)$ is differentiable at $x=0$
• D. $f\left(x\right)$ is not continuous at $x=0$.

Answer 1

Answer: (A), (C)

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

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

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

$=\frac{1}{2}\underset{y\to 0}{lim}\frac{\log (1-y)}{y}.\underset{x\to 0}{lim}{\left(\frac{\sin \left(x/2\right)}{x/2}\right)}^2\underset{x\to 0}{lim}\frac{x^2}{\log (1+x^2)}=-\frac{1}{2}$

where $y=1-\cos x$. So f is derivable at x=0 and hence also continuous.
Ans: A,C