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The value of the limit $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^{2}-x}+2 x\right)$ is
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$\frac{1}{4}$
Rationalise
$\lim _{x \rightarrow-\infty}\left(\left(\sqrt{4 x^{2}-x}+2 x\right) \times \frac{\sqrt{4 x^{2}-x}-2 x}{\sqrt{4 x^{2}-x}-2 x}\right)$
$\lim _{x \rightarrow-\infty}\left(\frac{-x}{|x| \sqrt{4-\frac{1}{x}}-2 x}\right)$ at $x \rightarrow-\infty \quad|x|=-x$
$\lim _{\mathrm{x} \rightarrow-\infty}\left(\frac{-\mathrm{x}}{-\mathrm{x} \sqrt{4-\frac{1}{\mathrm{x}}}-2 \mathrm{x}}\right)=\frac{1}{2+2}=\frac{1}{4}$
$\lim _{x \rightarrow-\infty}\left(\left(\sqrt{4 x^{2}-x}+2 x\right) \times \frac{\sqrt{4 x^{2}-x}-2 x}{\sqrt{4 x^{2}-x}-2 x}\right)$
$\lim _{x \rightarrow-\infty}\left(\frac{-x}{|x| \sqrt{4-\frac{1}{x}}-2 x}\right)$ at $x \rightarrow-\infty \quad|x|=-x$
$\lim _{\mathrm{x} \rightarrow-\infty}\left(\frac{-\mathrm{x}}{-\mathrm{x} \sqrt{4-\frac{1}{\mathrm{x}}}-2 \mathrm{x}}\right)=\frac{1}{2+2}=\frac{1}{4}$
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