If limit of function f(x)=(sin x-cos x)/cos 2x is l, choose the good answer: a)l=0;b)l=-1;c)=6;d)l=1/6;e)l=-square root 2/2

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We have to find the limit of f(x)=(sin x-cos x)/cos 2x for x--> 45 degrees.

We know that cos 2x = (cos x)^2 - (sin x )^2

lim x--> 0 [(sin x-cos x)/cos 2x]

=> lim x--> 0 [(sin x-cos x)/[(cos x)^2 - (sin x )^2]

=> lim x--> 0...

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We have to find the limit of f(x)=(sin x-cos x)/cos 2x for x--> 45 degrees.

We know that cos 2x = (cos x)^2 - (sin x )^2

lim x--> 0 [(sin x-cos x)/cos 2x]

=> lim x--> 0 [(sin x-cos x)/[(cos x)^2 - (sin x )^2]

=> lim x--> 0 [(sin x-cos x)/[(cos x - sin x)(cos x + sin x )]

cancel (sin x - cos x)

=> lim x--> 0 [-1/(cos x + sin x)]

substitute x = 0

=> -1/(cos x + sin x )

=> -1 / (1/sqrt 2 + 1/ sqrt 2)

=> -1/ (2/sqrt 2)

=> -1 / sqrt 2

=> -sqrt 2/2

l = -sqrt 2/2 or option e.

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