Question

limx tends to π/2 (1-sinx)/(π/2-x) ^2

Answer 1

After multiplying top and bottom by 1+sinx1+sin⁡x and rearranging, this is the same as

14limx→π/211+sinxcos2x(x−π/2)214limx→π/211+sin⁡xcos2⁡x(x−π/2)2

As x→π/2x→π/2, 1+sinx→21+sin⁡x→2. Now since cos(x+π/2)=−sinxcos⁡(x+π/2)=−sin⁡x, we can rewrite this as

14limx→π/2(11+sinx)limx→0(sinxx)2=1412(1)2

Hope it helps you
Please make me as brainliest