Question

Find the following limit:
$\underset{x\to 1}{lim}\frac{\sqrt[3]{3x}-1}{\sqrt{x}-1}.$

Answer 1

$=\underset{x\to 1}{lim}\frac{\sqrt[2]{2x}-1}{\sqrt{x}-1}$

$=\underset{x\to 1}{lim}\frac{\left[3\right]x-1}{\sqrt{x}-1}$

Put $x=t^3$

So ${lim}_{t\to 1}\frac{\sqrt[3]{3t^3}-1}{\sqrt{t^3}-1}$

$=\underset{t\to 1}{lim}\frac{t-1}{t^{3/2}-1^3}$

$=\underset{t\to 1}{lim}\frac{(t^{1/2}-1)(t^{1/2}+1)}{(t^{1/2}-1)(t^{1/4}+t^{1/2}+1)}$

$=\frac{2}{1+1+1}=\frac{2}{3}$.