# 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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### 2 Answers

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**

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.**