$$ \lim_\limits{n \to \infty}{\frac{10^n+n^2}{n!}} $$
Any hints on how to take on this limit? I do not know how to deal with the factorial in the denominator.
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1It there a typo. I don't see any factorial. Moreover is $n \to \infty$? – Anurag A Mar 02 '15 at 03:20
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Oops, typed in the wrong limit. Editing. – user3704351 Mar 02 '15 at 03:22
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1There are a host of tests (e.g., ratio test, root test, etc.) that can be used to find limits of these types. Which ones do you know? – Mark Viola Mar 02 '15 at 03:33
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Right now I know of the squeeze theorem. I am about to learn the root and ratio tests. – user3704351 Mar 02 '15 at 03:36
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$n!$ grows faster than any exponential function. – Akiva Weinberger Mar 02 '15 at 03:43
1 Answers
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Split the fraction up: ${10^n +n^2 \over n!}={10^n \over n!} + {n^2 \over n!}$
Note that:
${n^2 \over n!}={n \over n-1}{1 \over (n-2)!}$
Now $\lim_{n \to \infty}{1 \over (n-2)!}=0$ and $\lim_{n \to \infty}{n \over n-1}=1$, hence $\lim_{n \to \infty}{n^2 \over n!}=0$
Also, ${10^n \over n!}=\prod_{i=1}^{n}{10 \over i}=\left (\prod_{i=1}^{10}{10 \over i} \right ) \left (\prod_{i=11}^{n-1}{10 \over i} \right){10 \over n}≤\left (\prod_{i=1}^{10}{10 \over i} \right ){10 \over n}$ for $n≥11$.
Since $\left (\prod_{i=1}^{10}{10 \over i} \right )$ is a constant and ${10 \over n} \rightarrow 0$ as $n \rightarrow \infty$ we get $\lim_{n \to \infty}{10^n \over n!}=0$
Therefore: $\lim_{n \to \infty}{10^n +n^2 \over n!}=0+0=0$
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