Question
Evaluate the following limits:
$\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$

Answer

$\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}}{\sqrt{4\text{x}^2+1}-1}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\sqrt{4+\frac{1}{\text{x}^2}}-\frac{1}{\text{x}}}$
$=\frac{1}{\sqrt{4}-0}$
$=\frac12$

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