# How do you find the limit (sqrt(x)-1)/(root3(x)-1) as x->1?

Nov 9, 2016

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

#### Explanation:

What we can do here is fairly unintuitive. Recall that we can use the difference of cubes identity, or ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ to show that $x - 1 = \left(\sqrt[3]{x} - 1\right) \left(\sqrt[3]{{x}^{2}} + \sqrt[3]{x} + 1\right)$.

So, we can multiply the function by what could be considered its "cubic conjugate:"

${\lim}_{x \rightarrow 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x} - 1} = {\lim}_{x \rightarrow 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x} - 1} \cdot \frac{\sqrt[3]{{x}^{2}} + \sqrt[3]{x} + 1}{\sqrt[3]{{x}^{2}} + \sqrt[3]{x} + 1}$

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

We can also multiply by the conjugate of the term with the square root:

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

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

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

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

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

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

Nov 9, 2016

Answer: $\frac{3}{2}$

#### Explanation:

The $\frac{0}{0}$ form invokes the use of L'Hôpital's rule

$\frac{d \left({x}^{\frac{1}{2}} - 1\right)}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}}$

$\frac{d \left({x}^{\frac{1}{3}} - 1\right)}{\mathrm{dx}} = \frac{1}{3} {x}^{- \frac{2}{3}}$

${\lim}_{x \to 1} \frac{\frac{1}{2} {x}^{- \frac{1}{2}}}{\frac{1}{3} {x}^{- \frac{2}{3}}}$

${\lim}_{x \to 1} = \frac{3}{2} {x}^{\frac{1}{6}} = \frac{3}{2}$