Question

lim 2x tends to 0 sin²2x/2x² please dear friends solve it.I have my exams tomorrow​

Answer 1

Step-by-step explanation:

L hospital rule apply chseyali (0/0) or (infinity/infinity) form lo unnappudu

first step (1-cos2x)/2 = 2sin^x/2=sin^x

(1-cos4x)=2sin^2x

Answer 2

Answer:

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