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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4Divide numerator and denominator by $x$ and let $x\to \infty$ then. – martini Mar 27 '12 at 18:57
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yes, that is the simplest way to do it. the answer should be 1. – chango Mar 27 '12 at 18:58
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1@martini A brief explanation? Can you answer solving it? Thanks. – Garmen1778 Mar 27 '12 at 19:07
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1Mathlover did :) – martini Mar 27 '12 at 19:12
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possible duplicate of Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ – Aryabhata Mar 27 '12 at 19:34
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Add and Subtract $x$ from the terms and apply the result in the dupe: $(\sqrt{x^2 + 3x} - x) - (\sqrt{x^2 + x} - x)$. – Aryabhata Mar 27 '12 at 19:34
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Just curious could you substitute $x=1/y$ and solve for $\lim_{y\rightarrow0}$ ? – John Alexiou Mar 27 '12 at 20:46
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$\lim\limits_{ x\to \infty }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x } } } $
$\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3}{x}}+\sqrt{ 1+\frac{1}{x}}}}=\frac {2}{\large{\sqrt{ 1+0}+\sqrt{ 1+0}}}=1$
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