Q no 5 please provide a detailed solution 5 The value of limx→ /4 cos x - Maths - Limits and Derivatives

Question

Q no 5 please provide a detailed solution

5       T h e   v a l u e   o f   lim
x → π / 4   cos   x   +   sin   x
3   -   2 2 1   -   sin   2 x ,   i s
  ( a )     3 2          
                 
                 
                 
            ( b )     - 3 2
( c )     1 2            
                 
                 
                 
                 
  ( d )     - 1 2

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Brijendra Pal · Accepted Answer

Hi, limx→π4cosx+sinx3-221-sin2xafter putting limit it is 0/0 form so using L'hospital rule = limx→π4ddxcosx+sinx3-22ddx1-sin2x=limx→π43cosx+sinx2×-sinx+cosx-0-2cos2x=limx→π43cosx+sinx×cosx+sinx×-sinx+cosx-2cos2x=limx→π43cosx+sinx×cos2x-sin2x-2cos2x=limx→π43cosx+sinx×cos2x-2cos2x=limx→π43cosx+sinx-2=-3212+12=-3222=-32

Q no 5 please provide a detailed solution 5 . T h e v a l u e o f lim x → π / 4 cos x + sin x 3 - 2 2 1 - sin 2 x , i s ( a ) 3 2 ( b ) - 3 2 ( c ) 1 2 ( d ) - 1 2 ​

Q no 5 please provide a detailed solution

$5 . T h e v a l u e o f \lim_{x \to π / 4} \frac{{(\cos x + \sin x)}^{3} - 2 \sqrt{2}}{1 - \sin 2 x}, i s (a) \frac{\sqrt{3}}{2} (b) - \frac{3}{\sqrt{2}} (c) \frac{1}{2} (d) - \frac{1}{\sqrt{2}}$