Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2x+c_2}\right)$$
(provided it exists) by using the expression's conjugate, but this problem has me stumped.
I've considered using the conjugate
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
and a term like this one,
$$\sqrt{x^2+5x-2}+\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$
but that didn't seem to help simplify anything.
Edit: I stumbled across something at the last second that lets me use the conjugate approach. The expression can be rewritten as
$$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\\
\sqrt{x^2+5x-2}-2\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}+\sqrt{x^2+7x+5}\\
\left(\sqrt{x^2+5x-2}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)+\left(\sqrt{x^2+7x+5}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)$$
which approaches
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$

Rewrite it as:
$$\left(\sqrt{x^2+5x-2} - \left(x+\frac{5}{2}\right)\right) -\\ \left(\sqrt{4x^2-3x+7}-\left(2x-\frac{3}{4}
\right)\right) +\\
\left(\sqrt{x^2+7x+5}-\left(x+\frac{7}{2}\right)\right)+\left(\frac{5}{2}+\frac 34+\frac{7}2\right)$$
Or something like that. Each of the first three terms has limit zero...
