Find the limit as x approaches infinity of sin(2x)/x

The answer is 0. How can I show that this is an indeterminate form so I can use L'Hopital's? And even if I use L'Hopital's, it comes out to be 2cos(2x), which doesn't help if I replace x with infinity.

1 answer

You can't use L'Hopital's rule on this. Use the fact that sin 2x is bounded between -1 and +1 and the x in the denominator goes to infinity. The absolute value of the limit is equal to orless than the limit of 1/x as x-> infinity, which is zero
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