# Math Help - limits

1. ## limits

using a conjucate.. can someone show me how they would solve this? thanks.

lim x-> infinite [2x - squareroot(4x^2+1)]

2. Originally Posted by Legendsn3verdie
using a conjucate.. can someone show me how they would solve this? thanks.

lim x-> infinite 2x - squareroot(4x^2+1)
Multiplying by the conjugate we get

$\lim_{x\to\infty}(2x-\sqrt{4x^2+1})\cdot\frac{2x+\sqrt{4x^2+1}}{2x+\sqr t{4x^2+1}}$

Using the fact that $(f(x)-g(x))(f(x)+g(x))=f(x)^2-g(x)^2$ we get

$\lim_{x\to\infty}\frac{4x^2-(4x^2+1)}{2x+\sqrt{4x^2+12}}=\lim_{x\to\infty}\fra c{-1}{2x+\sqrt{4x^2+1}}=0$