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 lớn limit of 2sin(2x)/1. 2sin(2x)/1 as x goes khổng lồ 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 bởi you use?

Thank you so much.

The easiest use of the squeeze theorem for \$lim_x opminftyfracsin f(x)x\$ is \$-frac1lefracsin f(x)xlefrac1\$, so the limit is \$0\$.

If you are taking \$x o 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 o infty\$.

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