What is the limit for {[x + sin3x]/[2x + sinx]} when x tend to 0

lim_(x->0) {(x+sin(3 x))/(2 x+sin(x))}
Using the continuity of {x} at x = 4/3 write lim_(x->0) {(x+sin(3 x))/(2 x+sin(x))} as {lim_(x->0) (x+sin(3 x))/(2 x+sin(x))}:
= {lim_(x->0) (x+sin(3 x))/(2 x+sin(x))}
Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) (x+sin(3 x))/(2 x+sin(x)) = lim_(x->0) (( d(x+sin(3 x)))/( dx))/(( d(2 x+sin(x)))/( dx)):
= {lim_(x->0) (3 cos(3 x)+1)/(cos(x)+2)}
The limit of a quotient is the quotient of the limits:
= {(lim_(x->0) (3 cos(3 x)+1))/(lim_(x->0) (cos(x)+2))}
The limit of a sum is the sum of the limits:
= {(lim_(x->0) (3 cos(3 x)+1))/(lim_(x->0) cos(x)+2)}
The limit of cos(x) as x approaches 0 is 1:
= {1/3 (lim_(x->0) (3 cos(3 x)+1))}
The limit of a sum is the sum of the limits:
= {1/3 (3 (lim_(x->0) cos(3 x))+1)}
Using the continuity of cos(x) at x = 0 write lim_(x->0) cos(3 x) as cos(lim_(x->0) 3 x):
= {1/3 (3 cos(lim_(x->0) 3 x)+1)}
Factor out constants:
= {1/3 (3 cos(3 (lim_(x->0) x))+1)}
The limit of x as x approaches 0 is 0:
= {4/3}

1) Dividing numerator and denominator by x, the given expression =

= [1 + sin(3x)/x]/[2 + sin(x)/x]

2) = [1 + 3sin(3x)/(3x)]/[2 + sin(x)/x]

3) As limit x --> 0, both sin(3x)/(3x) and sin(x)/x = 1 [By trigonometry limit theorm]

4) Hence, the above limit = (1 + 3)/(2 + 1) = 4/3

EDIT:

Alternatively we can solve this by another method also, using sin(x) expansion.

sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

==> sin(3x) = 3x - (27x^3)/3! + (243x^5)/5! - ....

Hence, [x + sin(3x)]/[2x + sin(x)] =
= {x + 3x - (27x^3)/3! + (243x^5)/5! - ..}/[2x + x - (x^3)/3! + (x^5)/5! - ..]

As limit x --> 0, x^3 and all its higher powers become zero; So the it reduces to

= (4x + 0)/(3x + 0) = 4x/3x = 4/3

Lets call the limit L. Then we have:

L = lim x->0 {[x + sin3x]/[2x + sinx]}

Pull out x from the numerator and denominator. For ease of typing, I shall leave out the "lim x->0"; please understand that it is implicitly present.

So we have:
L = {x*[1 + (sin3x/x)]}/{x*[2 + (sinx/x)]}

This then becomes:
L = {[1 + (sin3x/x)]}/{[2 + (sinx/x)]}

Now we know that: lim x->0 [{sin(nx)}/(nx)] = 1, n= integer

Using that we have:
L = {[1 + 3*(sin3x/(3x))]}/{[2 + (sinx/x)]}
L = (1+3)/(2+1) = 4/3

Hence the limit value is 4/3

{[x+sin3x]/[2x+sinx] while x-> 0 then sinx also can be taken as just x

then it is [x+3x]/[2x+x]
=4x/3x
=4/3