lim as x --> 0 $\displaystyle [sin^2 (2x)]/x^2 $
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Use L'hospital rule
Originally Posted by roshanhero Use L'hospital rule i know the answer is 4 but i don't know how to get there...or how to use that rule
Can u differentiate sin^2(x) and x^2
Originally Posted by roshanhero Can u differentiate sin^2(x) and x^2 actually no i get it now. i have to multiply top and bottom by 2. that results in a special limit for sin. (2sin2x/2x)^2 =(2)^2 =4
yeap.............
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