Question

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

• A. -1
• B. 1
• C. 2
• D. None of these

Answer 1

The correct option is C

2

${lim}_{x\to \infty }\frac{\sqrt{1+x^4}+(1+x^2)}{x^2}$
$={lim}_{x\to \infty }\frac{\sqrt{x^4(\frac{1}{x^4}+(1+x^2\left)\right)}}{x^2}$
$={lim}_{x\to \infty }\frac{x^2\sqrt{\frac{1}{x^4}+1}(1+x^2)}{x^2}$
$={lim}_{x\to \infty }\frac{(\sqrt{\frac{1}{x^4}+1}\frac{1}{x^2}+1)}{x^2}$
$=\sqrt{0+1}+0+1=1+1$
$=2$