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Math Help - calculating the limit

  1. #1
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    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?
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  2. #2
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    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}}}
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  3. #3
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    Quote Originally Posted by bcahmel View Post
    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.
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  4. #4
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    thank you both very much!
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