Limit problem with roots

Question

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.

Answer 1

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 $$ a^9-b^9=(a-b)(a^8+a^7b+a^6b^2+a^5b^3+a^4b^4+a^3b^5+a^2b^6+ab^7+b^8) $$

Answer 2

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}