Q.
Lim (x-->pi) ((1-sin(x/2)) /(cos(x/2) * (cos (x/4) -sin (x/4)))?
Don't use L'hopitals Rule
Asked by sanjeev,
18 Feb '12 12:20 pm
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Answers (1)
1.
Lim (x) ((1-sin(x/2)) /(cos(x/2) * (cos (x/4) -sin (x/4)))
= lim (x) [(1-sin(x/2)*(1 + sin(x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(1-sin^2 (x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos^2 (x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos (x/2)] /[(1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos^2 (x/4) - sin^2 (x/4)] /[(1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos (x/4) + sin (x/4)] /[(1 + sin(x/2)]
= (1/2 + 1/2) / (1 + 1)
= 1/2.
Answered by anju chauhan, 18 Feb '12 12:24 pm
= lim (x) [(1-sin(x/2)*(1 + sin(x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(1-sin^2 (x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos^2 (x/2)] /[(cos(x/2) * (1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos (x/2)] /[(1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos^2 (x/4) - sin^2 (x/4)] /[(1 + sin(x/2) * (cos (x/4) -sin (x/4))]
= lim (x) [(cos (x/4) + sin (x/4)] /[(1 + sin(x/2)]
= (1/2 + 1/2) / (1 + 1)
= 1/2.
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