Verified answer

(tan(x) - sin(x)) / (x - x * cos(4x))

Remember this one: sin(x)/x goes to 1 as x goes to 0

sin(x) * (1/cos(x) - 1) / (x * (1 - cos(4x))) =>

(sin(x)/x) * ((1 - cos(x)) / cos(x)) / (1 - cos(4x)) =>

(sin(x)/x) * (1 - cos(x)) / (cos(x) * (1 - cos(4x)))

Let's focus on (1 - cos(x)) / (1 - cos(4x))

(1 - cos(x)) / (1 - cos(4x)) =>

(1 - cos(x)) / (1 - (cos(2x)^2 - sin(2x)^2)) =>

(1 - cos(x)) / (1 - cos(2x)^2 + sin(2x)^2) =>

(1 - cos(x)) / (sin(2x)^2 + sin(2x)^2) =>

(1 - cos(x)) / (2 * sin(2x)^2) =>

(1 - cos(x)) / (2 * (2 * sin(x) * cos(x))^2) =>

(1 - cos(x)) / (2 * 4 * sin(x)^2 * cos(x)^2) =>

(1 - cos(x)) / (8 * cos(x)^2 * (1 - cos(x)^2)) =>

(1 - cos(x)) / (8 * cos(x)^2 * (1 - cos(x)) * (1 + cos(x))) =>

1 / (8 * cos(x)^2 * (1 + cos(x)))

Now we have:

(sin(x)/x) / (cos(x) * 8 * cos(x)^2 * (1 + cos(x)))

x goes to 0

1 / (8 * cos(0)^3 * (1 + cos(0))) =>

1 / (8 * 1^3 * (1 + 1)) =>

1 / (8 * 2) =>

1/16

Answer

I agree with the 2nd responder's decision to factor out the quantity [sin(x)]/x, whose limit as x -> 0 is well known to be 1. That gives me:

lim x-->0 of [sec(x) - 1] / [1 - cos(4x)],

and that I can attack with L'Hopital's Rule:

lim x-->0 of [sec(x)tan(x)] / [4*sin(4x)] ... use L'Hopital's Rule again...

= lim x-->0 of [sec^3(x) + sec(x)tan^2(x)] / [16*cos(4x)]

= [1 + 0] / [16 * 1] = 1/16.

So I agree with cidyah's answer, but I believe this is a much easier way to get it.

lim x-->0 (tan x - sin x) /(x- x cos 4x)

substituting x=0, we have 0/0

Apply L'Hopital's rule

differentiate the numerator and denominator

= lim x-->0 (sec^2 x - cos x) / ( 1- (cos 4x + x (-4 sin 4x))

= lim x-->0 (sec^2 x - cos x) / ( 1 -cos 4x + 4x sin 4x)

again

d/dx ( sec^2 x - cos x) = 2 sec x sec x tan x - (-sin x) = 2 sec^2 x tan x + sin x

d/dx ( 1- cos 4x + 4x sin 4x) = (0 - (-4 sin 4x) + 4 sin 4x + 4x (4 cos 4x)) = 8 sin 4x+16x cos 4x)

= lim x-->0 ( 2 sec^2 x tan x + sin x) / (8 sin 4x + 16 x cos 4x)

again

d/dx ( 2 sec^2 x tan x + sin x) = 2 (2 sec x sec x tan x tan x + 2 sec^2 x sec^2 x + cos x)

=4 sec^2 x tan^2 x + 2 sec^4 x +cos x

d/dx ( 8 sin 4x + 16 x cos 4x) = 32 cos 4x + 16 cos 4x + 16 x ( -4 sin 4x)

= 48 cos 4x -64 x sin 4x

= lim x-->0 ( 4 sec^2 x tan^2 x + 2 sec^4 x + cos x) / (48 cos 4x -64 x sin 4x)

= lim x-->0 ( 0 + 2 + 1) / ( 48 - 0)

= 3/48

= 1/16