Solve the limit of fraction (tanx + sin2x)/ln(x+1)

You did not specified what is the accumulation
point.

We'll suppose that x approaches to
0.

We'll solve the limit, creating remarcable
limits.

lim tan x/x = 1, for
x->0

lim sin x/x = 1, for
x->0

lim ln(1+x)^(1/x) = ln e =
1

Now, we'll divide the numerator and denominator of the
fraction by x:

lim (tanx + sin2x)/ln(x+1) = lim[ (tanx)/x +
(sin2x)/x]/ln(x+1)/x

lim[ (tanx)/x + (sin2x)/x]/ln(x+1)/x =
lim[(tanx)/x + lim(sin2x)/x]/lim ln(x+1)/x

We'll solve lim
(sin2x)/x = lim 2*sin 2x/2x = 2lim sin 2x/2x = 2

lim (tanx
+ sin2x)/ln(x+1) = (1 + 2)/ln e = 3/1 = 3

The
limit of the given fractio, for x->0, is: lim (tanx + sin2x)/ln(x+1) =
3.