Question

$\underset{x\to \infty }{lim}\frac{(1-2+3-4+5-6+\dots -2n)}{\sqrt{(n^2+1)}+\sqrt{(4n^2-1)}}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\because 1-2+3-4+5-6+\dots \dots -2n=(-1)+(-1)+(-1)+\dots \dots ntimes=-n\therefore \underset{x\to \infty }{lim}\frac{(1-2+3-4+5-6+\dots -2n)}{\sqrt{(n^2+1)}+\sqrt{(4n^2-1)}}=\underset{x\to \infty }{lim}\frac{(-n)}{\sqrt{(n^2+1)}+\sqrt{(4n^2-1)}}=\underset{x\to \infty }{lim}\frac{1}{\sqrt{(1+\frac{1}{n^2})+\sqrt{(4-\frac{1}{n^2})}}}=\frac{-1}{\sqrt{1+0}+\sqrt{4-0}}=\frac{-1}{1+2}=-\frac{1}{3}$