Freedom without limits is not freedom

We multiply the expression in the limit by its "conjugate":

$\frac {\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} }{\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} }$

The limit becomes:

$\begin{aligned} & & \displaystyle \lim_{x \to \infty} \frac { (x^4 + ax^3 + 3x^2 + bx + 2) - (x^4 + 2x^3 - cx^2 + 3x - d)}{ \sqrt{x^4 + ax^3 + 3x^2 + bx + 2} + \sqrt{x^4 + 2x^3 - cx^2 + 3x - d} } \\ & = & \displaystyle \lim_{x \to \infty} \frac { x^3 (a-2) + x^2 (c+3) + x(b-3) + (d+2) } { x^2 \left ( \sqrt{1 + \frac {a}{x} + \frac {3}{x^2} + \frac {b}{x^3} + \frac {2}{x^4} } + \sqrt{1 + \frac { 2}{x} - \frac {c}{x^2} + \frac {3}{x^3} - \frac {d}{x^4} } \right ) } \\ \end{aligned}$

Focusing on the denominator, for $x \to \infty$ , the denominator $\to 2x^2$

If $a \ne 2$ , the the limit becomes $\frac {O(x^3)}{O(x^2)} \to \infty$ , thus $a = 2$

Now the expression of the limit is in the of $\frac {O(x^2)}{O(x^2)}$ , with the ratio of coefficient of $x^2$ in the numerator and denominator as $\frac {c+3}{2}$ , which equals to the given value $4$ , thus $c = 5$ , and so

$\large a - \frac {10}{c} = 2 - \frac {10} 5 = 0 \Rightarrow \left (b - \frac {4d}{5} \right ) \left (a - \frac {10}{c} \right) = 0$

Hence, our answer is simply $a + c = 2 + 5 = \boxed{7}$

using binomial expansion for rational index and neglecting higher degree terms one can easily evaluate this limit