# How do you find the limit lim (root4(x+1)-root4x)x^(3/4) as x->oo?

Feb 18, 2017

$\frac{1}{4}$

#### Explanation:

${\lim}_{x \to \infty} \left(\sqrt[4]{x + 1} - \sqrt[4]{x}\right) {x}^{\frac{3}{4}}$

= lim_(x to oo) x^(1/4)(root4(1+1/x))-1)x^(3/4)

$= {\lim}_{x \to \infty} x \left(\sqrt[4]{1 + \frac{1}{x}} - 1\right)$

By Biniomial Expansion:

$= {\lim}_{x \to \infty} x \left(\left(1 + \frac{1}{4 x} + O {\left(\frac{1}{x}\right)}^{2}\right) - 1\right)$

$= {\lim}_{x \to \infty} \frac{1}{4} + O {\left(\frac{1}{x}\right)}^{1} = \frac{1}{4}$

Feb 18, 2017

$\frac{1}{4}$
$\sqrt[4]{x + 1} - \sqrt[4]{x} = \frac{\sqrt[4]{{\left(x + 1\right)}^{2}} - \sqrt[4]{{x}^{2}}}{\sqrt[4]{x + 1} + \sqrt[4]{x}} = \frac{\sqrt[4]{{\left(x + 1\right)}^{4}} - \sqrt[4]{{x}^{4}}}{\left(\sqrt[4]{x + 1} + \sqrt[4]{x}\right) \left(\sqrt[4]{{\left(x + 1\right)}^{2}} + \sqrt[4]{{x}^{2}}\right)} = \frac{1}{\left(\sqrt[4]{x + 1} + \sqrt[4]{x}\right) \left(\sqrt[4]{{\left(x + 1\right)}^{2}} + \sqrt[4]{{x}^{2}}\right)} = \frac{1}{{x}^{\frac{3}{4}} \left(\sqrt[4]{1 + \frac{1}{x}} + \sqrt[4]{1}\right) \left(\sqrt[4]{{\left(1 + \frac{1}{x}\right)}^{2}} + \sqrt[4]{1}\right)}$
${\lim}_{x \to \infty} \left(\sqrt[4]{x + 1} - \sqrt[4]{x}\right) {x}^{\frac{3}{4}} = \frac{1}{4}$