Question

The value of $\underset{x\to -\infty }{lim}(x+\sqrt{x^2+2x})$ is

• A. $1$
• B. $-1$
• C. $-2$
• D. $\text{Does not exist}$

Answer 1

The correct option is B $-1$

$\underset{x\to -\infty }{lim}(x+\sqrt{x^2+2x})$
$=\underset{x\to -\infty }{lim}\frac{x^2-x^2-2x}{x-\sqrt{x^2+2x}}$

$=\underset{x\to -\infty }{lim}\frac{-2x}{x+x\sqrt{1+\frac{2}{x}}}[\because x<0\Rightarrow \sqrt{x^2}=-x]$

$=\underset{x\to -\infty }{lim}\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$
$=-1$