Let f:R→[0,∞] be such that limx→3 f(x)exists and limx→3 (f(x))2−4|x−3|=1 Then limx→3 f(x) equal

Since, $lim_{x \to 3} \frac{(f (x))^{2} - 4}{\sqrt{| x - 3 |}} = 1$ So

for $ϵ = 1$ there is $δ > 0$ such that

$0 < \frac{(f (x)^{2} - 9)}{\sqrt{| x - 5 |}} < 2 whenever 0 < | x - 3 | < δ$

$\Rightarrow 0 < f (x) - 2 < 2 \frac{\sqrt{| x - 3 |}}{f (x) + 2}$ whenever $0 < | x - 3 | < δ$

Since the range of $f$ is $[0, \infty]$ so $lim_{x \to 3} (f (x) + 2) \neq 0$ and exists. Thus

$0 \leq lim_{x \to 3} (f (x) - 2) \leq 0 \Rightarrow lim_{x \to 3} f (x) = 2$

Let $f : R \to [0, \infty]$ be such that $lim_{x \to 3} f (x)$
exists and $lim_{x \to 3} \frac{(f (x))^{2} - 4}{\sqrt{| x - 3 |}} = 1$ Then $lim_{x \to 3} f (x)$ equal