Question

Calculate the following limits.
$\underset{x\to 64}{lim}\frac{\sqrt{x}-8}{\sqrt[3]{3x}-4}.$

Answer 1

$\underset{x\to 64}{lim}\frac{\sqrt{x}-8}{\sqrt[3]{3x}-4}\left[\frac{0}{0}form\right]$

Apply L'hospital Rule : -

$\underset{x\to 64}{lim}\frac{\frac{1}{2\sqrt{x}}}{\frac{1}{3(x{)}^{2/3}}}=\underset{x\to 64}{lim}\frac{3(x{)}^{2/3}}{2\sqrt{x}}$

$=\underset{x\to 64}{lim}\frac{3}{2}\left(x^{\frac{1}{6}}\right)$

Now, Apply the limit

$=\frac{3}{2}\times 2=3$

$[\because (64{)}^{\frac{1}{6}}=2]$.