$\lim_{n\to \infty} (1+n+\cos n) ^\frac{1}{2n+n \sin n}$

While in class, we were proving a limit problem using the Squeeze Theorem, but when I was reviewing my notes, I came up with a problem,,

The first question was to prove that $$\lim_{n\to \infty}(1+n)^\frac{1}{n}=1$$

Okay, this was easy.

The next question was to use the limit proven above to evaluate the following limit: $$\lim_{n\to \infty}(1+n+n\cos n)^\frac{1}{2n+n\sin n}$$

In my notes, this was written;

$$1\leq(1+n+n\cos n)^\frac{1}{2n+n\sin n} \leq (1+2n+n\sin n)^\frac{1}{2n+n\sin n}$$

And since

$$\lim_{n\to \infty}(1+2n+n\sin n)^\frac{1}{2n+n\sin n}=1$$

Therefore by Squeeze Theorem, $$\lim_{n\to \infty}(1+n+n\cos n)^\frac{1}{2n+n\sin n}=1$$

My question is that the inequality doesn't seem to make sense. Is the inequality correct? Does it only hold for very large $n$ or something?

Then how would I evaluate this limit by using the first limit equation?

For all $n,$ $1\le 1+n + n\cos n \le 3n$ and $1/(2n + n\sin n) \le 1/n.$ Thus $$1 \le (1+n + n\cos n)^{1/(2n + n\sin n)}\le (3n)^{1/n} \to 1.$$
By the squeeze theorem, the limit is $1.$
What do you get if you put $n=2k +k\sin k$ in the first "easy" limit? Also, can you approximate $|\cos n-\sin n|$?
• If I put $n=2k+k\sin k$ in the first limit, I would get 1. – zxcvber Apr 10 '16 at 17:15