Limit of this expression [closed]

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Can you help me solve this limite? I tried to solve it with the expression $$U_n+_1\over U_n$$, but I got 0 * infinite in the end. $$U_n$$ is the expresion of the limit.

$$\lim_{n\to\infty}\frac{n! + 2^n log(n)}{3n! + n^2}$$

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2 Answers

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Hints

1. Consider$$ \frac{2^n\log n}{n!}<\frac{2^n}{(n-1)!} $$because $\log n<n$; now prove that $\dfrac{2^n}{(n-1)!}<\dfrac{1}{n}$ for $n>a$ (determine such $a$).

2. Prove that $\dfrac{n^2}{n!}<\dfrac{1}{n}$ for $n>b$ (determine such $b$).

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$\dfrac{n!+2^{n}\log{n}}{(3n)!+n^{2}} $~$\dfrac{n!(1+\frac{2^{n}\log{n}}{n!})}{(3n)!(1+o(1/n))}$~$\dfrac{n^{n}e^{3n}}{e^{n}(3n)^{3n} } + o(\frac{1}{n}) \to 0$, because of $n^{n}/n^{3n} $ goes to $0$ with $n\to \infty$

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