Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0

First, we'll substitute x by 0 into the
function:

lim ln(1+x)/(sinx+sin3x) = ln 1/(sin 0 + sin 0) =
0/0

We've get an
indetermination.

Since the trigonometric functions from
denominator are matching, we'll transform the sum into a
product:

sinx+sin3x = 2[sin
(x+3x)/2][cos(x-3x)/2]

sinx+sin3x = 2sin
2x*cos(-x)

Since the cosine function is even, we'll put
cos(-x) = cos x

sinx+sin3x = 2sin 2x*cos
x

We'll re-write the
limit:

lim ln(1+x)/(sinx+sin3x) = lim ln(1+x)/2sin 2x*cos
x

We'll re-arrange the terms, crating remarcable
limits:

(1/2)*lim ln(1+x)/sin 2x*lim (1/cos
x)

But lim (1/cos x) = 1/cos 0 =
1

(1/2)*lim ln(1+x)/sin 2x= (1/2)lim [ln(1+x)/x]*(2x/sin
2x)*(x/2x)

(1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x) =
(1/2)*1*1*lim (x/2x)

(1/2)lim (x/2x) =
1/4

For
x->0, lim ln(1+x)/(sinx+sin3x) =
1/4.