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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?

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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 $1.$

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What do you get if you put $n=2k +k\sin k$ in the first "easy" limit? Also, can you approximate $|\cos n-\sin n|$?

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  • $\begingroup$ If I put $n=2k+k\sin k$ in the first limit, I would get 1. $\endgroup$ – zxcvber Apr 10 '16 at 17:15
  • $\begingroup$ Do you mean approximation by using Taylor Series? $\endgroup$ – zxcvber Apr 10 '16 at 17:15
  • $\begingroup$ Do you see that you can use it to approximate the last expression of the double inequality? $\endgroup$ – MikeNerent93 Apr 10 '16 at 17:28

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