Question

If $a=\underset{x\to \infty }{lim}\left(\sqrt{x+\sqrt{x-}}\sqrt{x}\right)$ and $b=\underset{x\to \infty }{lim}(x-\sqrt{x+x^2})$ then

• A. $a+b=1$
• B. $a+b=0$
• C. $a=b$
• D. $a+b=2$

Answer 1

The correct option is B $a+b=0$

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

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

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

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