Question

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

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

$f:R\to [0,\infty ]\underset{x\to 5}{lim}\left(f\right(x){)}^2\frac{-9}{\sqrt{|x-5|}}=0$

if $x\to 5^{+}|x-5|=(x-5)$

then $\underset{x\to 5^{+}}{lim}\frac{\to 0}{\to 0}$ from

to bring this lim to exact $0$

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

$f\left(x\right)=\pm 3$

$f\left(x\right)=3$ [because $R\to [0,\infty ]]$

$\underset{x\to 0}{lim}f\left(x\right)=3$