$$20.\quad \lim_{x\to\infty}\frac{-6}{5x\sqrt[3]x} = -\frac65\lim_{x\to\infty}\frac1{x^{4/3}}= -\frac65\cdot 0 = 0$$

(Original scan of problem)

I cant figure out how to resolve this problem.

I would say that denominator tends to infinite and limit of -6 / infinite is 0. However the book seems to follow another way. Can you explain please?

  • $\begingroup$ The denominator grows without bound while the numerator remains fixed at $-6$. Hence the limit is zero. $\endgroup$ – user61527 Sep 23 '13 at 1:41
  • $\begingroup$ What you say is true and the book has just taken out $-\frac{6}{5}$ out of the limits because it's just a constant. $\endgroup$ – Sudarsan Sep 23 '13 at 1:42
  • $\begingroup$ But I am NOT happy with the approach the book did for Q21 !!! Frankly speaking, quite upset... $\endgroup$ – imranfat Sep 23 '13 at 1:44
  • 2
    $\begingroup$ @Sudarsan: You don't even need L'Hopital. Divide top and bottom by $x$. $\endgroup$ – Michael Albanese Sep 23 '13 at 1:47
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    $\begingroup$ @Jorge: Just to be clear, the word you should be using here is infinity, not infinite. Infinity is a noun, infinite is an adjective. $\endgroup$ – Michael Albanese Sep 23 '13 at 1:49

The book does exactly what you said, except for the fact that they first take out the constant $-\dfrac{6}{5}$.

Regarding your comments: One of the index laws is that $x^ax^b = x^{a+b}$. Together with the fact that for any positive integer $n$, $x^{\frac{1}{n}} = \sqrt[n]{x}$, you can get the desired denominator.

  • $\begingroup$ I would greatly appreciate a step by step resolution sorry its just I still cant get it. $\endgroup$ – JorgeeFG Sep 23 '13 at 1:48
  • $\begingroup$ What is confusing me is that I tried to reach X 4/3 and didnt make it. $\endgroup$ – JorgeeFG Sep 23 '13 at 1:54
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    $\begingroup$ $x x^{\frac{1}{3}} =x^1 x^{\frac{1}{3}} = x^{\frac{3}{3}}x^{\frac{1}{3}} = x^{\frac{3}{3}+\frac{1}{3}}=x^{\frac{4}{3}}$. $\endgroup$ – marty cohen Sep 23 '13 at 2:33

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