# How do you find the limit (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) as x->oo?

Aug 19, 2017

$\infty$

#### Explanation:

The greatest power of $x$ here is $1$, and is in the numerator. A handy rule of thumb is that then the numerator will outgrow the denominator, so the limit is $\infty$.

Another way to approach this is to factor the greatest power from the numerator and denominator:

lim_(xrarroo)(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4))=lim_(xrarroo)(x(1+x^(-1/2)+x^(-2/3)))/(x^(2/3)(1+x^(-5/12))

$= {\lim}_{x \rightarrow \infty} \frac{{x}^{\frac{1}{3}} \left(1 + {x}^{- \frac{1}{2}} + {x}^{- \frac{2}{3}}\right)}{1 + {x}^{- \frac{5}{12}}}$

$= {\lim}_{x \rightarrow \infty} {x}^{\frac{1}{3}} \cdot {\lim}_{x \rightarrow \infty} \frac{1 + {x}^{- \frac{1}{2}} + {x}^{- \frac{2}{3}}}{1 + {x}^{- \frac{5}{12}}}$

Note that for $p > 0$, ${\lim}_{x \rightarrow \infty} {x}^{-} p = 0$, so this limit becomes:

$= {\lim}_{x \rightarrow \infty} {x}^{\frac{1}{3}} \cdot \frac{1 + 0 + 0}{1 + 0}$

$= {\lim}_{x \rightarrow \infty} {x}^{\frac{1}{3}}$

$= \infty$

Aug 19, 2017

A very slightly different approach.

#### Explanation:

${\lim}_{x \rightarrow \infty} \frac{x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}}{{x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}}$ has initial form, $\frac{\infty}{\infty}$ which is indeterminate.

We will write a quotient that is equivalent, but whose denominator does not go to infinity.

Factor out of both the top and bottom, the greatest power of $x$ in the bottom. That is, factor out ${x}^{\frac{2}{3}}$.

$\frac{x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}}{{x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}} = \frac{x + {x}^{\frac{6}{12}} + {x}^{\frac{4}{12}}}{{x}^{\frac{8}{12}} + {x}^{\frac{3}{12}}}$

 = (x^(8/12)(x^(1/3)+1/x^(2/12)+1/x^(4/12)))/(x^(8/12)(1+1/x^(9/12))

$= \frac{{x}^{\frac{1}{2}} + \frac{1}{x} ^ \left(\frac{1}{6}\right) + \frac{1}{x} ^ \left(\frac{1}{3}\right)}{1 + {x}^{\frac{3}{4}}}$

${\lim}_{x \rightarrow \infty} \frac{{x}^{\frac{1}{2}} + \frac{1}{x} ^ \left(\frac{1}{6}\right) + \frac{1}{x} ^ \left(\frac{1}{3}\right)}{1 + {x}^{\frac{3}{4}}} = \frac{\infty + 0 + 0}{1 + 0} = \infty$