plz help me solve this question...
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Anonymous:
it should be 5/2
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check: as x→π/4
4√2 - (cos x + sin x)^5 becomes 0
1 - sin(2x) becomes 0
use L'Hopital's rule then .
= lim [-5(cos x + sin x)^4 (-sin x + cos x)] / [-2 cos(2x)]
= lim [5(cos x + sin x)^4 (-sin x + cos x)] / [2 {cos^2(x) - sin^2(x)}]
= lim [5(cos x + sin x)^4] [2 (cos x + sin x)]
= lim (5/2) (cos x + sin x)^3
= (5/2) (2√2)
= 5√2 .
4√2 - (cos x + sin x)^5 becomes 0
1 - sin(2x) becomes 0
use L'Hopital's rule then .
= lim [-5(cos x + sin x)^4 (-sin x + cos x)] / [-2 cos(2x)]
= lim [5(cos x + sin x)^4 (-sin x + cos x)] / [2 {cos^2(x) - sin^2(x)}]
= lim [5(cos x + sin x)^4] [2 (cos x + sin x)]
= lim (5/2) (cos x + sin x)^3
= (5/2) (2√2)
= 5√2 .
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