What is the limit as x=>0 of sin(x)/x+tan(x) ?

It is not clear if the denominator of the ratio is just x
or is (x+tan x).

This thing must be specified with the help
of the brackets.

If the denominator of the ratio is
x:

lim [sin(x)/x+tan(x)] = lim [sin(x)/x] + lim
tan(x)

lim [sin(x)/x] is an elementary limit and the result
is1.

We'll calculate lim tan
x.

We'll substitute x by the value
0.

lim tan x = tan 0 = 0

So,
the limit is:

lim [sin(x)/x+tan(x)] = 1 + 0 =
1

Now, if the denominator is
(x+tan x), we'll calculate the limit:

lim
[sin(x)/(x+tanx)]  = sin 0/(x + tan 0)

lim
[sin(x)/(x+tanx)]  = 0/0

"0/0" is an
indetermination

We'll use l'hospital rule. We'll
differentiate separately numerator and denominator.

(sinx)'
= cos x

(x+tanx)' = 1 + 1/(cos
x)^2

lim [sin(x)/(x+tanx)]  = lim
(sinx)'/(x+tanx)'

lim (sinx)'/(x+tanx)' = lim cos x/[1 +
1/(cos x)^2]

We'll substitute x by
0:

lim cos x/[1 + 1/(cos x)^2] =cos 0/[1 + 1/(cos
0)^2]

cos 0/[1 + 1/(cos 0)^2] = 1/(1+1) =
1/2

lim
[sin(x)/(x+tanx)]  =
1/2