Question

$\displaystyle \lim_{x \to -\infty}$$\frac{2x-1}{\sqrt{x^{2}+2x+1}}=$

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$\displaystyle \lim_{x \to -\infty}$$\frac{2x-1}{\sqrt{x^{2}+2x+1}}$
Put $x =-1/y$ as $x \to -\infty$, $y \to 0$
$=\displaystyle \lim_{y \to 0}$$\frac{-\frac{2}{y}-1}{\sqrt{\frac{1}{y^{2}}-\frac{2}{y}+1}}$
$=\displaystyle \lim_{y \to 0}$$\frac{-2-y}{\sqrt{1-2y+y^{2}}}$
$=-2$.