How to get rid of a cubic root in a logarithmic limit? The formula: 
$$\lim_{n \to \infty}\dfrac{\log (1 - n + n^2)}{\log (1 + n + n^{10})^{1/3}}.$$
Thanks for any advice!
 A: Hint:
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( 1-n+n^{2}\right) }{\frac{1}{3}
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) }.
\end{eqnarray*}
ADDED. Now that there is already a full answer I complete mine. Using the
rule $\log a^{r}=r\log a$, as commented by David Mitra, and $\log ab=\log
a+\log b$, manipulate algebrically the fraction and rewrite it as
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2\log n+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\times 10\log
n+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2+\frac{\log \left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+%
\frac{1}{3}\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{\log n}}.
\end{eqnarray*}
We thus have 
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\frac{\log \left( 1-n+n^{2}\right) }{\log \left(
1+n+n^{10}\right) ^{1/3}} &=&\lim_{n\rightarrow \infty }\frac{2+\frac{\log
\left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\frac{\log
\left( 1/n^{10}+1/n^{9}+1\right) }{\log n}} \\
&=&\frac{2+\displaystyle\lim_{n\rightarrow \infty }\frac{\log \left(
1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\displaystyle%
\lim_{n\rightarrow \infty }\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{
\log n}} \\
&=&\frac{2+\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty }\log n}}{%
\frac{10}{3}+\frac{1}{3}\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty
}\log n}} \\
&=&\frac{2+0}{\frac{10}{3}+\frac{1}{3}\times 0}=\frac{3}{5}.
\end{eqnarray*}
A: Solution with help of L'Hôpital's rule.
L'Hôpital's rule tells us, that if we have real valued functions $f$ and $g$ and assumed to be differentiable on an open interval with endpoint $a$, and additionally $g'(x)=0$ on the interval. It is also assumed that 
$$\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=L$$
 (Let $a$ and $L$ be extended real numbers)
If $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\infty $ , then 
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=L$$
I'll leave you check, that all condition are satisfied for your two function,
$$f(n)=\log (1 - n + n^2)$$ and,
$$g(n)=\log (1 + n + n^{10})^{1/3}=\frac{1}{3}\log (1 + n + n^{10}).$$
Now, let us find the derivatives: $f'(n)$ and $g'(n)$:
$$f'(n)=\frac{(1 - n + n^2)'}{1 - n + n^2}=\frac{2n-1}{1 - n + n^2}$$
$$g'(n)=\frac{1}{3}\frac{(1 + n + n^{10})'}{1 + n + n^{10}}=\frac{1}{3}\frac{(1 + 10n^{9})}{1 + n + n^{10}}$$
Now,
$$\frac{f'(n)}{g'(n)}=\frac{\frac{2n-1}{1 - n + n^2}}{\frac{1}{3}\frac{(1 + 10n^{9})}{1 + n + n^{10}}}=3\frac{(2n-1)(1 + n + n^{10})}{(1 - n + n^2)(1 + 10n^{9})}=\frac{2n^{11}+...}{10n^{11}+...}$$
The limit of last expression is,
$$\lim_{n\to\infty}\frac{f'(n)}{g'(n)}=\lim_{n\to\infty}3\frac{2n^{11}+...}{10n^{11}+...}=3\frac{2}{10}=\frac{3}{5}$$
Thus,
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac{3}{5}$$
