Let f: R → 0, ∞ be such that displaystyle limx→5 f(x) exists and displaystyle limx→5 ((f(x))2-9/√|x-5|) = 0 Then displaystyle limx→5 f(x) equals :

Question

Let f: R → 0, ∞ be such that displaystyle limx→5 f(x) exists and displaystyle limx→5 ((f(x))2-9/√|x-5|) = 0 Then displaystyle limx→5 f(x) equals : (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

ℓ imx→5 ((f(x)2)-9/√|x-5|) = 0 ℓ imx→ 5 [(f(x))2-9] = 0 ℓ imx→ 5 f(x) = 3

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

0

1

2

3

$ℓ i m_{x \to 5} \frac{( f ( x ) ^{2} ) - 9}{∣ x - 5 ∣} = 0$
$ℓ i m_{x \to 5} [(f (x))^{2} - 9] = 0$
$ℓ i m_{x \to 5} f (x) = 3$

Q. Let f:R→0,∞ be such that x→5lim​f(x) exists and x→5lim​∣x−5∣​(f(x))2−9​=0 Then x→5lim​f(x) equals :

0

1

2

3

ℓimx→5​∣x−5∣​(f(x)2)−9​=0 ℓimx→5​[(f(x))2−9]=0 ℓimx→5​f(x)=3

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

$ℓ i m_{x \to 5} \frac{( f ( x ) ^{2} ) - 9}{∣ x - 5 ∣} = 0$
$ℓ i m_{x \to 5} [(f (x))^{2} - 9] = 0$
$ℓ i m_{x \to 5} f (x) = 3$