How do you find lim (x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) as x->oo using l'Hospital's Rule?

Jun 11, 2017

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

Explanation:

You do not really need l'Hospital's rule as:

(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (x^(2/3) (x^(1-2/3) +x^(1/2-2/3) +x ^(1/3-2/3))) / (x^(2/3)( 1 + x^(1/4-2/3))

(x+x^(1/2)+x^(1/3))/(x^(2/3)+x^(1/4)) = (cancel(x^(2/3)) (x^(1/3) +x^(-1/6) +x ^(-1/3))) / (cancel(x^(2/3))( 1 + x^(-5/12))

$\frac{x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}}{{x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}} = \frac{{x}^{\frac{1}{3}} + {x}^{- \frac{1}{6}} + {x}^{- \frac{1}{3}}}{1 + {x}^{- \frac{5}{12}}}$

so as:

${\lim}_{x \to \infty} {x}^{\frac{1}{3}} + {x}^{- \frac{1}{6}} + {x}^{- \frac{1}{3}} = + \infty$

${\lim}_{x \to \infty} 1 + {x}^{- \frac{5}{12}} = 1$

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

Anyway, as the limit is in the form $\frac{\infty}{\infty}$ we can apply l'Hospital's rule:

${\lim}_{x \to \infty} \frac{x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}}{{x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \left(x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}\right)}{\frac{d}{\mathrm{dx}} \left({x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}\right)}$

${\lim}_{x \to \infty} \frac{x + {x}^{\frac{1}{2}} + {x}^{\frac{1}{3}}}{{x}^{\frac{2}{3}} + {x}^{\frac{1}{4}}} = {\lim}_{x \to \infty} \frac{1 + \frac{1}{2} {x}^{- \frac{1}{2}} + \frac{1}{3} {x}^{- \frac{1}{3}}}{\frac{2}{3} {x}^{- \frac{1}{3}} + \frac{1}{4} {x}^{- \frac{3}{4}}} = \frac{1}{0} = + \infty$