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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