IIT JAM
Deepak patra Asked a Question
April 4, 2021 11:15 pmpts 30 pts
  • IIT JAM
  • Mathematics (MA)

How lim sin²2x = 1 ,at x=0 by l'hospital rule 2cos(2x) in y=-lim- =-0.5 lim sin? (2x) =-0.5 x0 sin(2x).4.co sec (2x)cot(2x) x0 y = e-0.5

how lim sin²2x = 1 ,at x=0 By L'Hospital Rule 2cos(2x) In y=-lim- =-0.5 lim sin? (2x) =-0.5 x0 Sin(2x).4.co sec (2x)cot(2x) X0 y = e-0.5
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  • Alka gupta thankyou
    here some misprint.....limit x->0 sin^2(2x) is equal to 0 so limit should be e^0 = 1
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