# 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!

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Perhaps you want to use $\log a^r=r\log a$. – David Mitra Oct 17 '12 at 21:30
Thank you. I feel bad for not seeing it. – Grant Oct 17 '12 at 21:34
For some basic information about writing math at this site see e.g. here, here, here and here. – Américo Tavares Oct 17 '12 at 21:37

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*}

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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,

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

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