Let f : R→[0, ∞) be such that limx→5 f(x) exists and limx→5 (f(x))2−9|x−5|=0. Then, limx→5 f(x) equals

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

A
1
B
2
C
3
D
0

Solution:

$\begin{aligned} lim_{x \to 5} \frac{(f (x))^{2} - 9}{\sqrt{| x - 5 |}} = 0 \\ \Rightarrow lim_{x \to 5} (f (x))^{2} - 9 = 0 \end{aligned}$

$\Rightarrow$ $l^{2} - 9 = 0,$ where $\lim_{x \to 5} f (x) = l$

$\Rightarrow l = \pm 3$

$\Rightarrow l = 3$ $[∴ f (x) \geq 0$ for all $x \in R]$

$\Rightarrow \lim_{x \to 5} f (x) = 3$

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