Question

Lim
x Tends to pi/2
1-sinx/(2x-pi)√2​

Answer 1

Step-by-step explanation:

The best thing is to put x=π2+h. Now observe that as x→π/2, h→0. Also we get 2x−π=2h. Then your quantity becomes

limx→π/21−sin(x)(2x−π)2=limh→01−sin(π2+h)4h2=limh→01−cos(h)4h2=14⋅limh→02⋅sin2h2h2=14⋅limh→02⋅sin2h24⋅(h2)2

Answer 2

Answer:

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