What is the limit of the function sin x/(1-2sin^2x) -cos x/(2cos^2x-1) - sin2x/cosx, if x approaches to pi/4?
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We need to find the value of lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/2*(cos x)^2 - 1) - sin 2x / cos x.
Substituting x = pi/4, gives us an indeterminate value.
lim x--> pi/4 [sin x/(1- 2*(sin x)^2) - cos x/(2*(cos x)^2 - 1) - sin 2x / cos x
use (1- 2*(sin x)^2) = 2*(cos x)^2 - 1 = cos 2x
=> lim x--> pi/4 [ sin x/cos 2x - cos x/cos 2x - sin 2x/cos x]
=> lim x--> pi/4 [ (sin x - cos x)/cos 2x - sin 2x/cos x]
=> lim x--> pi/4 [ (sin x - cos x)/( (cos x)^2 - (sin x)^2) - sin 2x/cos x]
=> lim x--> pi/4 [ -1/(cos x + sin x) - sin 2x/cos x]
sin pi/4 = cos pi/4 = 1/sqrt 2
substituting x = pi/4
=> -1/((1/sqrt 2) + (1/sqrt 2)) - 1/(1/sqrt 2)
=> -1/sqrt 2 - sqrt 2
=> -3/sqrt 2
The required limit is -3/sqrt 2
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We recognize the double angle identities at the denominators of the first 2 terms.
We 'll re-write the denominators of th ratio as:
sin x/(cos 2x) -cos x/(cos 2x) - sin2x/cosx
But cos 2x = (cos x)^2 - (sin x)^2
We'll re-write the difference of squares as a product:
(cos x)^2 - (sin x)^2 = (cos x - sin x)(cos x + sin x)
We'll also re-write the numerator of the 3rd term of the function:
sin 2x = 2sin x*cos x
We'll re-write the function:
f(x) = (sin x-cos x)/(cos x - sin x)(cos x + sin x) - 2sin x*cos x/cos x
We'll simplify and we'll get:
f(x) = -1/(cos x + sin x) - 2sin x
Now, we'll take limit both sides:
lim f(x) = lim [-1/(cos x + sin x)] - 2 lim sin x
lim f(x) = -1/lim (cos x + sin x) - 2 lim sin x
We'll substitute x by pi/4:
lim f(x) = -1/(cos pi/4 + sin pi/4) - 2 sin pi/4
lim f(x) = -1/(sqrt2/2 + sqrt2/2) - 2sqrt2/2
lim f(x) = -2/2sqrt2 - 2sqrt2/2
lim f(x) = (-2 - 4)/2sqrt2
lim f(x) = -6/2sqrt2
lim f(x) = -3/sqrt2 = -3sqrt2/2
The limit of the given function, if x approaches to pi/4 is lim f(x) = -3sqrt2/2.
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