Question

Prove that :
$\underset{x\to 1}{lim}\frac{x^4-\sqrt{x}}{\sqrt{x}-1}=7$

Answer 1

L.H.S

$\underset{x\to 1}{lim}\frac{x^4-\sqrt{x}}{\sqrt{x}-1}$

This is $\frac{0}{0}$ form.

So, apply L-hospital rule,

$\underset{x\to 1}{lim}\frac{4x^3-\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}-0}$

$\underset{x\to 1}{lim}\frac{4x^3-\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$

$\Rightarrow \frac{4\times 1^3-\frac{1}{2\sqrt{1}}}{\frac{1}{2\sqrt{1}}}$

$\Rightarrow \frac{4-\frac{1}{2}}{\frac{1}{2}}$

$\Rightarrow \frac{\frac{7}{2}}{\frac{1}{2}}$

$\Rightarrow 7$

Hence, proved.