ns looked at the equipment of sin(2x)/x as x philosophies infinity (https://www.stillproud.orgway.com/popular-problems/Calculus/569134).

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Also, if you usage the L"hopital ascendancy instead of squeeze theorem for sin(2x)/x you acquire it is equal to border of 2sin(2x)/1. 2sin(2x)/1 as x goes to infinity is undefind ! therefore squeeze theorem states the initial limit is 0 while the l Hoptial ascendancy says the initial limit is undefined. Which rule do girlfriend use?

Thank girlfriend so much.


The easiest use the the squeeze theorem for $lim_x opminftyfracsin f(x)x$ is $-frac1xlefracsin f(x)xlefrac1$, therefore the limit is $0$.


If you space taking $x o infty$ girlfriend don"t have to worry around the instance where $x$ is negative.

You cannot use l"Hopital"s rule since the numerator $sin(2x)$ does not have a limit as $x o infty$.


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limit of $sinleft(fracpix ight)$ together $x$ philosophies $0$ does not exist, with squeeze theorem

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