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I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$

I did the following:

$$\begin{align*} \lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}} =& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\ \\ =& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\ \\ =& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\ \\ =& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}} \end{align*}$$

Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?

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From the final expression alone, there's no reason to think you did something wrong. – Git Gud Dec 22 '13 at 11:17
as per book the result is supposed to be 1... not sure how to get a 1 out of this. – Chris Dec 22 '13 at 11:18
@Chris: I wish I could simplify the fractions in Maths as you did in the last. Sigh! – Babak S. Dec 22 '13 at 11:23
omg.. now I see what was wrong.. – Chris Dec 22 '13 at 11:30
up vote 5 down vote accepted

Your last expression is completely wrong. This question is not solved in your way.



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can you explain me how to get to that? – Chris Dec 22 '13 at 11:33
You can multiply $x^{-1/3}$ to both. note that $x^0=1.$ – mathlove Dec 22 '13 at 11:35
I think I have studied too much today :) It would have been the easiest way to solve that.. thanks – Chris Dec 22 '13 at 11:37
A good day! You'll know a lot if you face many kinds of questions like this. You're welcome. – mathlove Dec 22 '13 at 11:39

$$F=\lim_{x \to \infty} \frac{\sqrt[3]x - \sqrt[5]x}{\sqrt[3]x + \sqrt[5]x}=\lim_{x \to \infty} \frac{x^{\frac13} - x^{\frac15}}{x^{\frac13} + x^{\frac15}}$$

As lcm$(3,5)=15$ I will set $\displaystyle x^{\frac1{15}}=y\implies x^{\frac13}=y^5$ and $\displaystyle x^{\frac15}=y^3$

$$\implies F=\lim_{y\to\infty}\frac{y^5-y^3}{y^5+y^3}$$

Setting $\frac1y=h,$ $$F=\lim_{h\to0}\frac{(1-h^2)h^3}{(1+h^2)h^3}=\lim_{h\to0}\frac{1-h^2}{1+h^2}$$ the cancellation of $h$ is legal as $h\ne0$ as $h\to0$

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@Chris: Setting $x=y^{15}$ is a very useful kind of "trick." It transforms something you may not have good intuition about into something familiar. – André Nicolas Dec 22 '13 at 16:32

Your major mistake : $x^a + x^b \ne x^{a+b}$. This is very wrong and a common mistake.

Did you know that $\sqrt[n]{x} = x^{1/n}$? Therefore the intial expression can be written as : $$\frac{x^{1/3} - x^{1/5}}{x^{1/3} + x^{1/5}} = \frac{x^{1/3}(1 - x^{1/5-1/3})}{x^{1/3}(1+x^{1/5-1/3})} = \frac{1-x^{-2/15}}{1+x^{-2/15}}$$

What do you know about $x^{-a}$ ($a>0$) as $x\longrightarrow \infty$?

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$$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}=\lim_{x \to \infty} \frac{x^{1/3}-x^{1/5}}{x^{1/3} + x^{1/5}}=$$ $$=\lim_{x \to \infty} \frac{1-x^{1/5-1/3}}{1 + x^{1/5-1/3}}=\lim_{x \to \infty} \frac{1-x^{-2/15}}{1 + x^{-2/15}}=\lim_{x \to \infty} \frac{1-\frac{1}{x^{2/15}}}{1 + \frac{1}{x^{2/15}}}=1$$

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