# lim_n→->∞infty(1/3+n^1000/2^n)^n calculate

$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}}\right)}^{n}$calculate
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spornya1
If $f\left(n\right)=\left(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}}{\right)}^{n}$,
$\begin{array}{}g\left(n\right)& =\mathrm{ln}f\left(n\right)\\ & =n\mathrm{ln}\left(\frac{1}{3}+\frac{{n}^{1000}}{{2}^{n}}\right)\\ & =n\mathrm{ln}\left(\frac{1}{3}\left(1+\frac{3{n}^{1000}}{{2}^{n}}\right)\right)\\ & =n\left(\mathrm{ln}\left(1/3\right)+n\mathrm{ln}\left(1+\frac{3{n}^{1000}}{{2}^{n}}\right)\\ & =-n\mathrm{ln}\left(3\right)+n\mathrm{ln}\left(1+\frac{3{n}^{1000}}{{2}^{n}}\right)\end{array}$
Since $\frac{3{n}^{1000}}{{2}^{n}}\to 0$ and $\mathrm{ln}\left(1+x\right)\sim x$ for small x, $n\mathrm{ln}\left(1+\frac{3{n}^{1000}}{{2}^{n}}\right)\sim n\frac{3{n}^{1000}}{{2}^{n}}\to 0$ so $g\left(n\right)\sim -n\mathrm{ln}\left(3\right)=\mathrm{ln}\left({3}^{-n}\right)$ and $f\left(n\right)\sim 1/{3}^{n}\to 0$.
Used ln, not exp.