# If lim_(x rarr oo)((ax^2+bx+3)^(1/2) -2x)=7, then find a and b?

May 17, 2018

${\lim}_{x \rightarrow \infty} \left(\sqrt{4 {x}^{2} + 28 x + 3} - 2 x\right) = 7$

#### Explanation:

Complete the square underneath the radical to get:

${\lim}_{x \rightarrow \infty} \left(\sqrt{a} \sqrt{{\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2}}{4 a c} + \frac{3}{a}} - 2 x\right)$
$= {\lim}_{x \rightarrow \infty} \left(x \sqrt{a} + \frac{b}{2 \sqrt{a}} - 2 x\right) = 7$

Okay, so we have this equivalent form, and note that this is the equation of the line, and so for it to have a finite limit of seven, the $x$ terms above have to go to 0.

Thus $x \sqrt{a} - 2 x = 0 R i g h t a r r o w \left(x\right) \left(\sqrt{a} - 2\right) = 0 R i g h t a r r o w \sqrt{a} = 2 R i g h t a r r o w a = 4$

In this case our limit becomes

${\lim}_{x \rightarrow \infty} \left(x \sqrt{a} + \frac{b}{2 \sqrt{a}} - 2 x\right) = {\lim}_{x \rightarrow \infty} \left(x \sqrt{4} + \frac{b}{2 \sqrt{4}} - 2 x\right) = \frac{b}{4} = 7 R i g h t a r r o w b = 28$

So in fact this is satisfied only by $a = 4$ and $b = 28$