Question

Let $f:R\to \left[0,\infty \right)$ be such that $\underset{x\to 5}{lim}f\left(x\right)$ exists and $\underset{x\to 5}{lim}\frac{{\left(f\left(x\right)\right)}^2-9}{\sqrt{\left|x-5\right|}}=0.$ Then $\underset{x\to 5}{lim}f\left(x\right)$ equal

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

Given: $f:R\to [0,\infty ):{lim}_{x\to 5}f\left(x\right)$

Given: ${lim}_{x\to 5}\frac{{\left(f\left(x\right)\right)}^2-9}{\sqrt{|x-5|}}=0$

$\Rightarrow {lim}_{x\to 5}\frac{{\left(f\left(x\right)\right)}^2-9}{\sqrt{|x-5|}}=0$

We know that denominator cannot be zero$\Rightarrow \sqrt{|x-5|}\ne 0$

So for making ${lim}_{x\to 5}\frac{{\left(f\left(x\right)\right)}^2-9}{\sqrt{|x-5|}}$ equal to zero,

${lim}_{x\to 5}{\left[f\left(x\right)\right]}^2-9$ should be equal to zero.

${lim}_{x\to 5}{\left(f\left(x\right)\right)}^2-9=0$

$\Rightarrow {lim}_{x\to 5}{\left(f\left(x\right)\right)}^2=9$

$\Rightarrow {lim}_{x\to 5}f\left(x\right)=\pm 3$

$\Rightarrow {lim}_{x\to 5}f\left(x\right)\ne -3[\because f:R⟶(0,\infty )]$

$\Rightarrow {lim}_{x\to 5}f\left(x\right)=3$