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

We 'll re-write the denominator of th ratio
as:

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 re-write the
function:

f(x) = (sin x-cos x)/(cos x - sin x)(cos x + sin
x)

We'll simplify and we'll
get:

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

Now, we'll take limit both
sides:

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

lim f(x) = -1/lim (cos x + sin
x)

lim f(x) = -1/(cos pi/4 + sin
pi/4)

lim f(x) = -1/(sqrt2/2 +
sqrt2/2)

lim f(x) =
-1/2sqrt2/2

lim f(x) = -sqrt
2/2

Since the limit of the function is l,
then l = -sqrt 2/2, so the good answer is e).