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

Earn 10 points for answering

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.

Report abuse

Useful

(0)

Not Useful

(0)

Your vote on this answer has already been received