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$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{x^{2}+2 x+1}}$ is equal to
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The correct answer is:
$-2$
$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{x^{2}+2 x+1}}$
$=\lim _{y \rightarrow \infty} \frac{-2-\frac{1}{y}}{\sqrt{1-\frac{2}{y}+\frac{1}{y^{2}}}} \quad[$ put $x=-y, x \rightarrow-\infty]$
$=-\frac{2}{1}=-2$
$=\lim _{y \rightarrow \infty} \frac{-2-\frac{1}{y}}{\sqrt{1-\frac{2}{y}+\frac{1}{y^{2}}}} \quad[$ put $x=-y, x \rightarrow-\infty]$
$=-\frac{2}{1}=-2$
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