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

(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?

MJD
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JorgeeFG
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  • The denominator grows without bound while the numerator remains fixed at $-6$. Hence the limit is zero. –  Sep 23 '13 at 01:41
  • 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 01:42
  • But I am NOT happy with the approach the book did for Q21 !!! Frankly speaking, quite upset... – imranfat Sep 23 '13 at 01:44
  • @imranfat Ha ha, I guess L'hopital should have been the best description for that ! The book's answer seems to be quite arbitrary. – Sudarsan Sep 23 '13 at 01:46
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    @Sudarsan: You don't even need L'Hopital. Divide top and bottom by $x$. – Michael Albanese Sep 23 '13 at 01:47
  • Sudarsan and Michael, you are both correct and considering the 3 problems I suspect that they may not know L'Hospital yet, and so Michael's method would then be appropriate, but not the way how the book presented it. Jorge, take a note of Michael's remark when you move on to Q21 – imranfat Sep 23 '13 at 01:49
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    @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 01:49
  • I think the 21 is just taking the most representative X. – JorgeeFG Sep 23 '13 at 01:51
  • Yes Jorge, that is true. But the way how they "knocked off" the constants is not right. Dividing every term by x (which is the leading term as you mentioned) and then take the limit, is the appropriate way. Problems like this where x goes to infinity come back with radicals in the fraction and then taking off the constants will likely go wrong. – imranfat Sep 23 '13 at 01:58

1 Answers1

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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.

Michael Albanese
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