Question

If $f\left(x\right)=9^{-2/x^2},x\ne 0$ is continuous at $x=0,$ then the value of $f\left(0\right)$ is

• A. $0$
• B. $1$
• C. $9$
• D. $-1$

Answer 1

The correct option is A $0$

Given : $f\left(x\right)=9^{-2/x^2},x\ne 0$
At $x=0$,
$R.H.L.=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{9}^{-2/h^2}=0$
$(\because h\to 0,\frac{-2}{h^2}\to -\infty )$
Since $f\left(x\right)$ is continuous at $x=0$
$\Rightarrow R.H.L.=f\left(0\right)=0$
$\therefore f\left(0\right)=0$