 Carla Murphy

2021-12-27

Evaluating $\underset{n\to \mathrm{\infty }}{lim}{\left(1+\frac{\mathrm{sin}n}{5n+1}\right)}^{2n+3}$ (a ${1}^{\mathrm{\infty }}$ indeterminate form) karton

Take logarithm and write it as
$\frac{2}{5}\left(\left(5n+1+\frac{13}{2}\right)\mathrm{ln}\left(1+\frac{\mathrm{sin}n}{5n+1}\right)\right)$
the limit becomes
$=\frac{2}{5}\underset{n\to +\mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(1+\frac{\mathrm{sin}n}{5n+1}\right)}{\frac{\mathrm{sin}n}{5n+1}}\mathrm{sin}n$
$=\underset{n\to +\mathrm{\infty }}{lim}\mathrm{sin}n$
which does not exist.
Observe that
$\underset{n\to +\mathrm{\infty }}{lim}\frac{13}{5}\mathrm{ln}\left(1+\frac{\mathrm{sin}n}{5n+1}\right)=0$
So, your limit does not exist. user_27qwe

Define
$L:=\left(1+\frac{\mathrm{sin}n}{5n+1}{\right)}^{2n+3}$
and take the logarithm of both sides to bring the exponent down. Proceed from there. Vasquez

Using $1+rx<\left(1+x{\right)}^{r}, we have
$1+\frac{2n+3}{5n+1}\mathrm{sin}n<\left(1+\frac{\mathrm{sin}n}{5n+1}{\right)}^{2n+3}
For any n such that $\mathrm{sin}n>\frac{1}{2}$ (and there are infinitely many), we have
$1+\frac{2n+3}{5n+1}\mathrm{sin}n>1+\frac{2}{5}\cdot \frac{1}{2}=1.2$
For any n such that $\mathrm{sin}n<-\frac{1}{2}$ (also infinitely many), we have
$exp\left(\frac{2n+3}{5n+1}\mathrm{sin}n\right)
And thus the limit can't exist, as the value is infinitely often less than 0.9 and infinitely often greater than 1.2.

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