Question

Evaluate the limit : $\underset{x\to 1}{lim}\frac{\sqrt{x+8}}{\sqrt{x}}$

Answer 1

Using direct substitution method,

$\underset{x\to 1}{lim}\frac{\sqrt{x+8}}{\sqrt{x}}=\frac{\sqrt{1+8}}{\sqrt{1}}$

$=\frac{\sqrt{9}}{1}$

$=3$

Therefore,

$\underset{x\to 1}{lim}\frac{\sqrt{x+8}}{\sqrt{x}}=3$