Question

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

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $-\frac{1}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $-\frac{1}{3}$

$\frac{(1+3+5-----2n-1)-(2+4+6-----2n)}{\sqrt{n^2+1}+\sqrt{4n^2-1}}$

$=\frac{(1+2+3+4+5+6-----2n)-2(2+4+6-----2n)}{\sqrt{n^2+1}+\sqrt{4n^2-1}}$

$=\frac{\frac{2n(2n+1)}{2}-4\left(\frac{n(n+1)}{2}\right)}{\sqrt{n^2+1}+\sqrt{4n^2-1}}$

$=\frac{n(2n+1-2n-2)}{n(\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}})}=\frac{-1}{1+2}$

$=\frac{-1}{3}$