Question

The value of $\underset{x\to \infty }{lim}\frac{\sqrt{1+x^4}-(1+x^2)}{x^2}$ is

• A. -1
• B. 0
• C. 2
• D. none of these

Answer 1

The correct option is B 0

$\underset{x\to \infty }{lim}\frac{\sqrt{1+x^4}-(1+x^2)}{x^2}$
$=\underset{x\to \infty }{lim}\frac{\sqrt{1+x^4}-(1+x^2)}{x^2}.\frac{\sqrt{1+x^4}+(1+x^2)}{\sqrt{1+x^4}+(1+x^2)}$
$=\underset{x\to \infty }{lim}\frac{(1+x^4)-(1+x^2{)}^2}{x^2}.\frac{1}{\sqrt{1+x^4}+(1+x^2)}$
$=\underset{x\to \infty }{lim}\frac{-2}{\sqrt{1+x^4}+(1+x^2)}=\frac{-2}{\infty }=0$