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
$$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$

You can also use Taylor series. Start writing $$A=\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}$$ $$\frac Ax=\sqrt{1+\frac{5x-2}{x^2}}-2\sqrt{1-\frac{3x-7}{4x^2}}+\sqrt{1+\frac{7x+5}{x^2}}$$ Now, remember that, when $y$ is small compared to $1$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$.
So replace $y$ by $\frac{5x-2}{x^2}$ for the first radical, by $-\frac{3x-7}{4x^2}$ for the second radical and by $\frac{7x+5}{x^2}$ for the third radical. So, in the order, each radical write $$\sqrt{1+\frac{5x-2}{x^2}}=1+\frac{5}{2 x}-\frac{33}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ $$\sqrt{1-\frac{3x-7}{4x^2}}=1-\frac{3}{8 x}+\frac{103}{128 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ $$\sqrt{1+\frac{7x+5}{x^2}}=1+\frac{7}{2 x}-\frac{29}{8 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ All of the above makes $$\frac Ax=\frac{27}{4 x}-\frac{599}{64 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ that is to say $$A=\frac{27}{4}-\frac{599}{64 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ which gives  the limit but also tells how it is approched.
