Evaluating $\lim_{n\to\infty}\left(1 + \frac{\sin n}{5n + 1}\right)^{2n + 3}$ (a $1^\infty$ indeterminate form) My professor gave us this limit as part of homework:
$$
\lim_{n\to\infty}\left(1 + \frac{\sin n}{5n + 1}\right)^{2n + 3}
$$
I can see it's in the indeterminate form $1^{\infty}$, so my first thought was to write it in the form $\lim_{n\to0}\left(1 + n\right)^{\frac{1}{n}}$ since I know that's equal to $e$, so I tried that:
$$
\begin{align*}
\lim_{n\to\infty}\left(1 + \frac{\sin n}{5n + 1}\right)^{2n + 3} &= \\
\lim_{n\to\infty}\left(1 + \frac{\sin n}{5n + 1}\right)^{\left(2n+3\right)\cdot\left(\frac{1}{\frac{\sin n}{5n + 1}}\right)\cdot\left(\frac{\sin n}{5n + 1}\right)} &= \\
\lim_{n\to\infty}\left(\left(1 + \frac{\sin n}{5n + 1}\right)^{\frac{1}{\frac{\sin n}{5n + 1}}}\right)^{\frac{2n + 3}{5n + 1}\cdot\sin n} &=
\end{align*}
$$
Up to this point I was happy, 'cause I successfully wrote it in that form, but I got stuck when I tried to solve the limit of the outermost exponent, since it doesn't have a limit.
I apologize in advance for any grammatical or sentence structure errors, English is not my first language. Anyways, any feedback about this problem, like hints, where my error was, or literally anything, is extremely accepted!
 A: Using $1 + rx < (1 + x)^r < \exp(rx)$, we have
$$
1 + \frac{2n+3}{5n+1}\sin n < \left(1 + \frac{\sin n}{5n+1}\right)^{2n + 3} <  \exp\left(\frac{2n+3}{5n+1}\sin n\right).
$$
For any $n$ such that $\sin n > 1/2$ (and there are infinitely many), we have
$$
1 + \frac{2n+3}{5n+1}\sin n > 1 + \frac{2}{5}\cdot\frac{1}{2} = 1.2.
$$
For any $n$ such that $\sin n < -1/2$ (also infinitely many), we have
$$
\exp\left(\frac{2n+3}{5n+1}\sin n\right) < \exp\left(-\frac{2}{5}\cdot\frac{1}{2}\right) < 0.9.
$$
And thus the limit can't exist, as the value is infinitely often less than $0.9$ and infinitely often greater than $1.2$.
A: Define
$$L:=\left( 1+\frac{\sin n}{5n+1} \right)^{2n+3},$$
and take the logarithm of both sides to bring the exponent down. Proceed from there.
A: Take logarithm and write it as
$$\frac 25((5n+1+\frac{13}{2})\ln\Bigl(1+\frac{\sin(n)}{5n+1}\Bigr)$$
the limit becomes
$$=\frac 25\lim_{n\to+\infty}\frac{\ln\Bigl(1+\frac{\sin(n)}{5n+1}\Bigr)}{\frac{\sin(n)}{5n+1}}\sin(n)$$
$$=\lim_{n\to+\infty}\sin(n)$$
which does not exist.
Observe that
$$\lim_{n\to+\infty}\frac{13}{5}\ln(1+\frac{\sin(n)}{5n+1})=0$$
So, your limit does not exist.
A: $$\lim_{n\to\infty}\left(1 + \frac{\sin n}{5n + 1}\right)^{2n + 3} = \lim_{n\to\infty}\left[\left[\color{red}{\left(1 + \frac{\sin n}{5n + 1}\right)^{\frac{5n+1}{\sin n}}}\right]^{\color{green}{\frac{2n+3}{5n+1}}}\right]^{\sin n}$$
The limit of red part is $e$, green part is $\frac25$ and black part doesn't exist. So the overall limit oscillates between $e^{\frac25}$ and $e^{-\frac25}$ and doesn't exist.
