Question

$\underset{n\to \infty }{lim}\frac{1-2+3-4+5-6+.......2n}{\sqrt{n^2+1}+\sqrt{4n^2-1}}\text{is equal to:}$

• A. 1/3
• B. -1/3
• C. -1/5
• D. 1/5

Answer 1

The correct option is B

-1/3

$\underset{n\to \infty }{lim}\frac{1-2+3-4+5-6+.......2n}{\sqrt{n^2+1}+\sqrt{4n^2-1}}=\underset{n\to \infty }{lim}\frac{[1+3+5+7+.....+(2n-1\left)\right]-(2+4+6+.....+2n)}{\sqrt[n]{n1+\frac{1}{n^2}}+\sqrt[n]{n4-\frac{1}{n^2}}}=\underset{n\to \infty }{lim}\frac{\frac{n}{2}(2.1+(n-1\left).2\right)-\frac{n}{2}[2.2+(n-1\left).2\right]}{n[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}]}=\underset{n\to \infty }{lim}\frac{\frac{n}{2}.2n-\frac{n}{2}2(n+1)}{n[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}]}=\underset{n\to \infty }{lim}\frac{n^2-n^2-n}{n[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}]}=\underset{n\to \infty }{lim}\frac{-n}{n[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}]}=\underset{n\to \infty }{lim}\frac{-1}{[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}]}=\frac{-1}{1+2}=-\frac{1}{3}$