# How to find the limit $\lim\limits _{ x\to \infty } \left( \sqrt { x^2+3x } -\sqrt { x^2+x } \right)$?

When I try to do this type of indeterminations I reach to this point:

$\lim\limits_{ x\to \infty } \dfrac { 2x }{ \sqrt { x^ 2 +3x } +\sqrt { x^2 +x } }$

but I don't know how to continue. Thanks.

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Divide numerator and denominator by $x$ and let $x\to \infty$ then. – martini Mar 27 '12 at 18:57
yes, that is the simplest way to do it. the answer should be 1. – chango Mar 27 '12 at 18:58
@martini A brief explanation? Can you answer solving it? Thanks. – Garmen1778 Mar 27 '12 at 19:07
Mathlover did :) – martini Mar 27 '12 at 19:12
– Aryabhata Mar 27 '12 at 19:34

$\lim\limits_{ x\to \infty }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x } } }$
$\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3}{x}}+\sqrt{ 1+\frac{1}{x}}}}=\frac {2}{\large{\sqrt{ 1+0}+\sqrt{ 1+0}}}=1$