Calculating $ \lim_{n\to \infty} (1+\sin({1}/{n}))^{n}$ without L'Hopital or series expansions

Question

I am trying to calculate the following limit, without using the L'Hopital rule or series expansions:

lim (1+sin(1/n))^(n), n->infinity

I now that it is the same as:

lim (1+sin(n))^(1/n), n->0

But that's about all I know...

Any help?

Answer 1

$$\lim_{n\to\infty}\left(1+\sin\frac1n\right)^n$$

$$=\left[\lim_{n\to\infty}\left(1+\sin\frac1n\right)^\frac1{\sin\frac1n}\right]^{\lim_{n\to\infty}n\cdot\sin\frac1n}$$

Set $\sin\frac1n=1/m$ to get $\lim_{n\to\infty}\left(1+\sin\dfrac1n\right)^\frac1{\sin\frac1n}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

Finally for the exponent, $\lim_{n\to\infty} n\cdot\sin\dfrac1n=\lim_{h\to0}\dfrac{\sin h}h=1$

Answer 2

This is a duplicate. I copy below my answer

Using Taylor expansion we have $$\sin t=t (1+o(1))\text{ when }t\to 0.$$ At this point if $x$ goes to infinity, then $$\left(1+sin \frac{1}{x}\right)^x \sim \left(1+\frac{1}{x}\right)^x\to e.$$