Evaluate the following limit : Lt x tends 0 1-cos 2x / sin ^2 2x
Please explain the steps too?
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12
Hope that this will help. CHEERS!!
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abhi178:
check qn again
there's a question Lt x---> 0 1- cos mx/ 1- cosnx
in this question why do we need to multiply with m and n
Answered by
35
Lt x_0 {1 - cos2x}/sin^2(2x)
we know,
(1 - cos2x) = 2sin^2x use this here,
Lt x_0 {2sin^2 x }/sin^2 (2x)
we also know ,
sin2x = 2sinx.cosx use this here ,
Lt x_0 {2sin^2x }/{2sinx.cosx}^2
Lt x_0 {2sin^2x }/4sin^2x.cos^2x
Lt x_0 1/2cos^2x
now put x=0
= 1/2(1)^2 = 1/2
hence answer = 1/2
we know,
(1 - cos2x) = 2sin^2x use this here,
Lt x_0 {2sin^2 x }/sin^2 (2x)
we also know ,
sin2x = 2sinx.cosx use this here ,
Lt x_0 {2sin^2x }/{2sinx.cosx}^2
Lt x_0 {2sin^2x }/4sin^2x.cos^2x
Lt x_0 1/2cos^2x
now put x=0
= 1/2(1)^2 = 1/2
hence answer = 1/2
can a limit be 1/1
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