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!

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,
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac{3}{5}$$
