Question

find the limit of (sin(x))/x as x approaches infinity

Answer 1

I actually know the answer is zero because wolfram alpha and the solution guide both say it is. Both also say that it is bounded therefore it equals zero. That doesn't make sense to me because the only theorem I know of that says anything about the argument being bounded is the bounded monotonic sequence theorem. but that theorem says the argument has to be bounded AND monotonic. (sin(x))/x is not monotonic.

Answer 2

\[-1 \le \sin(x) \le 1\]\[\Rightarrow \dfrac{-1}{x} \le \dfrac{\sin(x)}{x} \le \dfrac{1}x{}\]\[\Rightarrow \lim_{x\to \infty} \dfrac{-1}{x} \le \lim_{x \to \infty} \dfrac{\sin(x)}{x} \le \lim_{x \to \infty}\dfrac{1}{x}\]\[\Rightarrow 0 \le \lim_{x \to \infty} \dfrac{\sin(x)}{x} \le 0\]\[\Rightarrow \lim_{x\to \infty}\dfrac{\sin(x)}x{} = 0\]

Answer 3

Ahhh the squeeze theorem I didn't even think of that thank you!

Answer 4

I am about to post another one that is giving me a hard time maybe you can help if you have a moment @ParthKohli

Answer 5

I don't know if I'd be able to help. I'm not that much of an expert but lol :-P