Question

if $f\left(x\right)=\underset{n\to \infty }{\lim }\left(\frac{tan\pi x^2+{\left(x+1\right)}^{n}sinx}{x^2+{\left(x+1\right)}^n}\right)$, then

• A. $f$ is continuous at $x=0$
• B. $f$ is differentiable at $x=0$
• C. $f$ is continuous but not differentiable at $x=0$
• D. None of these

Answer 1

Answer: (D)

$f\left(0\right)=\frac{0+1\times 0}{0+1}=0$
$\underset{x\to 0^{-}}{\lim }f(x)=$$\underset{x\to 0^{-}}{\lim }$$\underset{x\to \infty }{\lim }$$\frac{tan\pi x^2+{\left(x+1\right)}^{n}sinx}{x^2+{\left(x+1\right)}^n}$
$=\underset{x\to 0^{-}}{\lim }$$\frac{tan\pi x^2}{x^2}$ (If $x\to 0^{-},x+1<1$)
$=\pi \therefore LHL\ne f(0)$
$\therefore f(x)$ is not continuous at $x=0$ hence not differentiable also.