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$\lim_{n \to \infty } \frac{10^{n}+n^{20}}{n!} = 0$

 

Please explain how

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$\lim_{n \to \infty } \frac{10^{n}+n^{20}}{n!} =\lim_{n \to \infty } \frac{10^n}{n!} + \lim_{n \to \infty }\frac{n^{20}}{n!} $

Comparing $10^n$ and $n!$, take log both sides, we get $nlog(10)$ and $log(n!)$.

log(n!) = log(1) + log(2) + log(3) + ....... + log(n)
log(n!) <= log(n) + log(n) + log(n) + .... + n-times
log(n!) <= nlog(n)

So for large values of $n$ (say $\color{blue}{10^{100}}$) $nlog(n)$ grows faster than $nlog(10)$, thus $n!$ grows faster than $10^n$

So, $\lim_{n \to \infty } \frac{10^n}{n!} = 0$

Similarly, $n!$ grows faster than $n^{20}$, So, $\lim_{n \to \infty }\frac{n^{20}}{n!} = 0$

So, we can conclude that $\large \lim_{n \to \infty } \frac{10^{n}+n^{20}}{n!} =0$

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When n-> infinite,

10ⁿ = o( n! ),

n^(20) = o(n!).

Therefore their limit when tends to infinite becomes 0.

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