# calculating the limit

• Dec 13th 2010, 01:42 PM
bcahmel
calculating the limit
Problem: lim x--> infinity, w $x^{4/3}+4x^{1/3}$ all divided by $x^{5/4} -2x$.(sry couldn't figure out how to make it into a fraction).

From there I divided the numerator and the denominator by x^(5/4), and I ended up with $x^{1/12}+4x^{-11/12}$ divided by $1-2x^{-1/4}$.

I kind of confused on what I do from here to find the limit...Any help please?
• Dec 13th 2010, 02:54 PM
You don't need to go through all this. When working with rational expressions like this, just use the highest powers.

$\lim_{x->\infty}\frac{x^{\frac{4}{3}}+4x^{\frac{1}{3}}}{x^ {\frac{5}{4}}-2x}=\lim_{x->\infty}\frac{x^{\frac{4}{3}}}{x^{\frac{5}{4}}}$
• Dec 13th 2010, 04:58 PM
Prove It
Quote:

Originally Posted by bcahmel
Problem: lim x--> infinity, w $x^{4/3}+4x^{1/3}$ all divided by $x^{5/4} -2x$.(sry couldn't figure out how to make it into a fraction).

From there I divided the numerator and the denominator by x^(5/4), and I ended up with $x^{1/12}+4x^{-11/12}$ divided by $1-2x^{-1/4}$.

I kind of confused on what I do from here to find the limit...Any help please?

What you are doing is correct, but it's easiest to write each term as its own fraction...

$\displaystyle \lim_{x \to \infty}\frac{x^{\frac{4}{3}} + 4x^{\frac{1}{3}}}{x^{\frac{5}{4}} - 2x} = \lim_{x \to \infty}\frac{x^{\frac{1}{12}} + \frac{4}{x^{\frac{11}{12}}}}{ 1 - \frac{2}{x^{\frac{1}{4}}}}$.

All the terms that have $\displaystyle x$ in the denominator go to $\displaystyle 0$, while the $\displaystyle x^{\frac{1}{12}}$ still goes to $\displaystyle \infty$. So the limit is $\displaystyle \infty$.
• Dec 14th 2010, 12:36 PM
bcahmel
thank you both very much!