# What is lim_(x->0) (sqrt(x+1)-1)/(root(3)(x+1)-1) ?

Jan 14, 2018

${\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1} = \frac{3}{2}$

#### Explanation:

Let:

$t = \sqrt[6]{x + 1}$

Then:

${\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1} = {\lim}_{t \to 1} \frac{{t}^{3} - 1}{{t}^{2} - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = {\lim}_{t \to 1} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(t - 1\right)}}} \left({t}^{2} + t + 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(t - 1\right)}}} \left(t + 1\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = {\lim}_{t \to 1} \frac{{t}^{2} + t + 1}{t + 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = \frac{3}{2}$

Alternative method

Alternatively, we can use the binomial theorem to find:

${\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1} = {\lim}_{x \to 0} \frac{{\left(1 + x\right)}^{\frac{1}{2}} - 1}{{\left(1 + x\right)}^{\frac{1}{3}} - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = {\lim}_{x \to 0} \frac{\left(1 + \frac{1}{2} x + O \left({x}^{2}\right)\right) - 1}{\left(1 + \frac{1}{3} x + O \left({x}^{2}\right)\right) - 1}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = {\lim}_{x \to 0} \frac{\frac{1}{2} x + O \left({x}^{2}\right)}{\frac{1}{3} x + O \left({x}^{2}\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = {\lim}_{x \to 0} \frac{\frac{1}{2} + O \left(x\right)}{\frac{1}{3} + O \left(x\right)}$

$\textcolor{w h i t e}{{\lim}_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x + 1} - 1}} = \frac{3}{2}$