Question

$\underset{x\to 0}{lim}\frac{\sqrt{8-3x}+\sqrt{8+4x}}{\sqrt{2-3x}}$= ______

• A. 5
• B. 3
• C. 2
• D. 4

Answer 1

The correct option is D 4

Given,
$\underset{x\to 0}{lim}\frac{\sqrt{8-3x}+\sqrt{8+4x}}{\sqrt{2-3x}}$
Substituting $x=0$, we get
$\underset{x\to 0}{lim}\frac{\sqrt{8-3x}+\sqrt{8+4x}}{\sqrt{2-3x}}=\frac{\sqrt{8-3\left(0\right)}+\sqrt{8+4\left(0\right)}}{\sqrt{2-3\left(0\right)}}=\frac{2\sqrt{8}}{\sqrt{2}}=2\times \sqrt{4}=2\times 2=4$