Question

The value of the limit $\underset{x\to -\infty }{lim}(\sqrt{4x^2-x}+2x)$ is

• A. $-\infty$
• B. $-\frac{1}{4}$
• C. $0$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

$\underset{x\to -\infty }{lim}(\sqrt{4x^2-x}+2x)$
$=\underset{x\to -\infty }{lim}(\sqrt{4x^2-x}+2x)\times \frac{(\sqrt{4x^2-x}-2x)}{(\sqrt{4x^2-x}-2x)}$
$=\underset{x\to -\infty }{lim}\left(\frac{-x}{|x|\sqrt{4-\frac{1}{x}}-2x}\right)$

At $x\to -\infty ,|x|=-x$
$\underset{x\to -\infty }{lim}\left(\frac{-x}{-x\sqrt{4-\frac{1}{x}}-2x}\right)=\frac{1}{2+2}=\frac{1}{4}$