# Limits to infinite

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

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?

• The denominator grows without bound while the numerator remains fixed at $-6$. Hence the limit is zero. – user61527 Sep 23 '13 at 1:41
• 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. – Sudarsan Sep 23 '13 at 1:42
• But I am NOT happy with the approach the book did for Q21 !!! Frankly speaking, quite upset... – imranfat Sep 23 '13 at 1:44
• @Sudarsan: You don't even need L'Hopital. Divide top and bottom by $x$. – Michael Albanese Sep 23 '13 at 1:47
• @Jorge: Just to be clear, the word you should be using here is infinity, not infinite. Infinity is a noun, infinite is an adjective. – 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.
• $x x^{\frac{1}{3}} =x^1 x^{\frac{1}{3}} = x^{\frac{3}{3}}x^{\frac{1}{3}} = x^{\frac{3}{3}+\frac{1}{3}}=x^{\frac{4}{3}}$. – marty cohen Sep 23 '13 at 2:33