Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ 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?
 A: Your last expression is completely wrong. This question is not solved in your way. 
HINT:
$$\frac{x^{1/3}-x^{1/5}}{x^{1/3}+x^{1/5}}=\frac{1-x^{(1/5)-(1/3)}}{1+x^{(1/5)-(1/3)}}$$
A: $$F=\lim_{x \to \infty} \frac{\sqrt[3]x - \sqrt[5]x}{\sqrt[3]x + \sqrt[5]x}=\lim_{x \to \infty} \frac{x^{\frac13} - x^{\frac15}}{x^{\frac13} + x^{\frac15}}$$
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 $h\to0$
A: 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 $x\longrightarrow \infty$?
A: $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}=\lim_{x \to \infty} \frac{x^{1/3}-x^{1/5}}{x^{1/3} + x^{1/5}}=$$
$$=\lim_{x \to \infty} \frac{1-x^{1/5-1/3}}{1 + x^{1/5-1/3}}=\lim_{x \to \infty} \frac{1-x^{-2/15}}{1 + x^{-2/15}}=\lim_{x \to \infty} \frac{1-\frac{1}{x^{2/15}}}{1 + \frac{1}{x^{2/15}}}=1$$
