I looked at the solution of sin(2x)/x as x approaches infinity (https://www.nasaconstellation.comway.com/popular-problems/Calculus/569134).

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I understand -1

Also, if you use the L"hopital rule instead of squeeze theorem for sin(2x)/x you get it is equal to limit of 2sin(2x)/1. 2sin(2x)/1 as x goes to infinity is undefind ! So squeeze theorem says the original limit is 0 while the L Hoptial rule says the original limit is undefined. Which rule do you use?

Thank you so much.

The easiest use of the squeeze theorem for $\lim_{x\to\pm\infty}\frac{\sin f(x)}{x}$ is $-\frac{1}{|x|}\le\frac{\sin f(x)}{x}\le\frac{1}{|x|}$, so the limit is $0$.

If you are taking $x \to \infty$ you don"t have to worry about the case where $x$ is negative.

You cannot apply l"Hopital"s rule because the numerator $\sin(2x)$ does not have a limit as $x \to \infty$.

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Limit of $\sin\left(\frac{\pi}{x} \right)$ as $x$ approaches $0$ does not exist, with squeeze theorem