Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
2  
@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

4 Answers 4

up vote 4 down vote accepted

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

HINT:

$$\frac{x^{1/3}-x^{1/5}}{x^{1/3}+x^{1/5}}=\frac{1-x^{(1/5)-(1/3)}}{1+x^{(1/5)-(1/3)}}$$

share|improve this answer
    
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

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 $x\longrightarrow \infty$?

share|improve this answer

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

share|improve this answer

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

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 $h\to0$

share|improve this answer
1  
@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 Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.