<s>[INST] 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
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$
 [/INST] $\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right) = 27/4 $
i will use the binomial theorem in the form 
$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small+\cdots $  to each.
$$\sqrt{x^2+5x-2} = x + {5x - 2 \over 2x} + \cdots \\
-\sqrt{4x^2-3x+7} = -\left( 2x  + {-3x + 7 \over 4x} + \cdots \right) \\
\sqrt{x^2+7x+5} = x + {7x + 5 \over 2x} + \cdots \\
$$
adding the dominant terms the limit is $${5 \over 2} + {3 \over 4} + {7 \over 2} = {10 + 3 + 14\over 4} = {27 \over 4} $$
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