Limit involving square roots, more than two "rooted" terms

Question

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

Answer 1

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...

Answer 2

$\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 $${5 \over 2} + {3 \over 4} + {7 \over 2} = {10 + 3 + 14\over 4} = {27 \over 4} $$

Answer 3

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.

Answer 4

Hint: $\sqrt{ax^2+bx+c} = \dfrac{\sqrt{a+\dfrac{b}{x}+\dfrac{c}{x^2}}}{\dfrac{1}{x}} = \dfrac{\sqrt{a+by+cy^2}}{y}$,and use L'hospitale rule. Note $y \to 0$. Repeat this process for each term of the sum and simplify to a common fraction and proceed to L'hospitale rule.