# Find the limit?

## lim_(x rarr -1)(1+root(3)x)/(1+root(5)x

Apr 1, 2018

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = \frac{5}{3}$

#### Explanation:

Given: ${\lim}_{x \rightarrow - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}}$

Because the expression evaluated at the limit, yields the indeterminant form, $\frac{0}{0}$, one should use L'Hôpital's rule.

To apply L'Hôpital's rule, one differentiates the numberator and the denominator with respect to the dependent variable:

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = {\lim}_{x \to - 1} \frac{\frac{d \left(1 + \sqrt[3]{x}\right)}{\mathrm{dx}}}{\frac{d \left(1 + \sqrt[5]{x}\right)}{\mathrm{dx}}}$

Perform the differentiations:

lim_(x to -1)(1+root(3)x)/(1+root(5)x) = lim_(x to -1)(1/3x^(-2/3))/(1/5x^(-4/5)

Simplify

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = {\lim}_{x \to - 1} \frac{5}{3} {x}^{- \frac{2}{15}}$

The limit on the right can be evaluated at $x = - 1$:

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = \frac{5}{3}$

Apr 1, 2018

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = \frac{5}{3}$

#### Explanation:

The limit:

${\lim}_{x \to - 1} \frac{1 + {x}^{\frac{1}{3}}}{1 + {x}^{\frac{1}{5}}}$

is in the indeterminate form $\frac{0}{0}$ so we can use l'Hospital's rule:

${\lim}_{x \to - 1} \frac{1 + {x}^{\frac{1}{3}}}{1 + {x}^{\frac{1}{5}}} = {\lim}_{x \to - 1} \frac{\frac{d}{\mathrm{dx}} \left(1 + {x}^{\frac{1}{3}}\right)}{\frac{d}{\mathrm{dx}} \left(1 + {x}^{\frac{1}{5}}\right)}$

${\lim}_{x \to - 1} \frac{1 + {x}^{\frac{1}{3}}}{1 + {x}^{\frac{1}{5}}} = {\lim}_{x \to - 1} \frac{\frac{1}{3} {x}^{- \frac{2}{3}}}{\frac{1}{5} {x}^{- \frac{4}{5}}} = \frac{5}{3}$

graph{(1+root(3)x)/(1+root(5)x) [-2, 0, -1, 5]}

Apr 1, 2018

See below

#### Explanation:

As the root is odd we will handle only the real root.

Making $y = \sqrt[15]{x}$ we have

$\frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = \frac{1 + {y}^{3}}{1 + {y}^{5}} =$

$= \frac{\left(y - 1\right) \left({y}^{4} - {y}^{3} + {y}^{2} - y + 1\right)}{\left(y - 1\right) \left({y}^{2} - y + 1\right)} = \frac{{y}^{4} - {y}^{3} + {y}^{2} - y + 1}{{y}^{2} - y + 1}$

and finally

${\lim}_{x \to - 1} \frac{1 + \sqrt[3]{x}}{1 + \sqrt[5]{x}} = {\lim}_{y \to - 1} \frac{{y}^{4} - {y}^{3} + {y}^{2} - y + 1}{{y}^{2} - y + 1} = \frac{5}{3}$