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I am given that $\lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx=1$ need to find the value of $a,b \in \mathbb R$ such that the given limit is true. I was able to work the whole thing out, but I have a question about one step in my work. There is a lot of rough work because I simplify by using the rule of the difference of cubes, so here is a condensed part of my work: $$\begin{align} \lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx &=\lim_{x \to \infty} \frac{8x^3+ax^2-b^3x^3}{(\sqrt[3]{8x^3+ax^2})^2+bx\sqrt[3]{8x^3+ax^2}+b^2x^2} \\&= \lim_{x \to \infty} \frac{8+a\frac{1}{x}-b^3}{\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+b^2\frac{1}{x}} \\&=\frac{\lim_{x \to \infty}8+\lim_{x \to \infty}a\frac{1}{x}-\lim_{x \to \infty}b^3}{\lim_{x \to \infty}\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+\lim_{x \to \infty}b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+\lim_{x \to \infty}b^2\frac{1}{x}} \\&= \frac{8-b^3}{0+0+0} \\&= \frac{8-b^3}{0}\end{align}$$ Thus $8-n^3$ must also equal $0$ which implies that $b=2$. (This is the part I am unsure about. Is what I said true? If $b=2$ then this would give me an indeterminate form, but other than that I'm not sure if what I said holds, and if it does hold why does it hold?) Regardless of my uncertainty, I went on and using this assumption I found that $a=12$ in a similar manner, and when I check $\lim_{x \to \infty} \sqrt[3]{8x^3+12x^2}-2x$ it does equal $1$ .

Any help as to why/why not my assumption is correct? Thanks in advance!

(If anyone wants me to post the method as to how i got 12 for $a$, let me know and then I'll type it up).

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It looks good...until you write division by zero. I think I understand what you mean and I think you're right, yet try to avoid explicitly writing that: we mathematicians usually begin to pant and some of us get the rabbies when we see that thing... –  DonAntonio Apr 23 '13 at 3:08
    
@DonAntonio lol. That is where my confusion came about, I don't really know how to avoid it. –  user66807 Apr 23 '13 at 3:12
    
Stop your stuff two lines before you did, and argue that since the first limit exists and the denominator of the expression you reached in the last step is zero, then in the last step it must be that also the denominator has limit zero... –  DonAntonio Apr 23 '13 at 3:14
    
@DonAntonio you mean "also the numerator has limit zero," right? –  user66807 Apr 23 '13 at 3:18
    
Yes, of course. Thanks. –  DonAntonio Apr 23 '13 at 3:47

1 Answer 1

up vote 2 down vote accepted

Your assumption is correct; if $b\neq 2$, then $8-b^3\neq 0$, and hence the limit would either not exist or be infinite. But you know the limit is $1$.

A shorter way to do the first part is: $\sqrt[3]{8x^3+ax^2}-bx=x(\sqrt[3]{8+\frac{a}{x}}-b)$. The cube root approaches 2 as $x\rightarrow \infty$, so if $b\neq 2$, the product approaches $\pm\infty$ (not $1$ as in the hypothesis).

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Wow...your method really is so much shorter and simpler. I'm amazed. But I have a question now: If $8-b^3\neq 0$ then how can we be sure that the limit would either not exist or be infinite. I think I'm missing the concept. –  user66807 Apr 23 '13 at 3:08
    
$8-b^3$ is a nonzero constant in this case, while the denominator approaches 0. This cannot approach 1. If the denominator is always of the same sign, then the fraction will always be of the same sign, and hence approach either $+\infty$ or $-\infty$. However if the denominator changes signs, the limit doesn't exist. –  vadim123 Apr 23 '13 at 3:14
    
Okay, now I have another question. In your method the cube root approaches 2 as $x \to \infty$ but if $b=2$ then the "stuff" in the parenthesis will tend to $0$ and then we will have $\infty * 0$ won't we? –  user66807 Apr 23 '13 at 3:22
    
Correct. The $\infty \cdot 0$ form is indeterminate, so it might equal 1. Any other value for $b$ cannot give a limit of 1. Hence $b$ must be $2$. –  vadim123 Apr 23 '13 at 3:23
    
Ahh okay I understand. Thanks a lot for your help! –  user66807 Apr 23 '13 at 3:27

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