Lim x->0 [sqrt(4+x^2+3x^3)-2]/[2x^3+3x^2]

I don't want L'HOSPITAL RULE
And the correct answer is 1/12
but i can't find it; I repeated this exercise so many times and it give me the 8/3 answer

Try the old trick of multiplying top and bottom by the conjugate surd expression, i.e. by
√(4+x²+3x³) + 2
Then the top becomes
4+x²+3x³ - 4 = x²+3x³

while the bottom is (2x³+3x²)[√(4+x²+3x³) + 2]

Now divide both top and bottom by x²
and the top becomes 1+3x
and the bottom becomes
(2x+3)[√(4+x²+3x³) + 2]

Now as x approaches 0, this approaches
(1+3*0) / (2*0 + 3)[√(4+0+3*0) + 2]

= 1 / 3(2+2)
= 1/12