Limit problem with roots I'm struggling solving the following limit problem:
$$\lim_{x \to \infty} \left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)$$
At first I thought I could 
Multiply by:
$$\frac{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}$$
But that doesn't seem to take me anywhere closer to an answer.
Some help would be appreciated.
 A: Substitute $t=1/x$, so the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt[9]{1+t}-\sqrt[9]{1-t}}{t}
$$
Now apply l'Hôpital or Taylor. Surely the conjugate is not
$\sqrt[9]{1+t}+\sqrt[9]{1-t}$ that would be only for the square root.
Actually, this is the derivative at $0$ of the function
$$
f(t)=\sqrt[9]{1+t}-\sqrt[9]{1-t}
$$
and, since
$$
f'(t)=\frac{1}{9}(1+t)^{-8/9}+\frac{1}{9}(1-t)^{-8/9}
$$
we have
$$
f'(0)=\frac{2}{9}
$$
If you have to do it without derivatives, you have to use
$$
a^9-b^9=(a-b)(a^8+a^7b+a^6b^2+a^5b^3+a^4b^4+a^3b^5+a^2b^6+ab^7+b^8)
$$
A: Alternatively:
\begin{align}
&\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\\
=&\frac{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)\\
=&\frac{\left({x^9+x^8}\right)-\left({x^9-x^8}\right)}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
=&\frac{2x^8}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
\end{align}
Therefore,
\begin{align}
&\lim_{x\to\infty} \sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\\
=&\lim_{x\to\infty}\frac{2x^8}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
=&\lim_{x\to\infty}\frac{2x^8}{\sum_{k=0}^8x^{8-k}x^k}\\
=&\lim_{x\to\infty}\frac{2x^8}{9\cdot x^8}\\
=&\frac{2}{9}
\end{align}
