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?

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.