Question

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

• A. $-1$
• B. $1$
• C. $2$
• D. $noneofthese$

Answer 1

Answer: (C)

$2$

We have,

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

Now,

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

$\Rightarrow {lim}_{x\to \infty }\frac{x^2[\sqrt{(\frac{1}{x^4}+1)}+(\frac{1}{x^2}+1)]}{x^2}$

$\Rightarrow {lim}_{x\to \infty }\sqrt{(\frac{1}{x^4}+1)}+(\frac{1}{x^2}+1)$

Taking limit and we get,

$\sqrt{(\frac{1}{\infty }+1)}+(\frac{1}{\infty }+1)$

$\Rightarrow \sqrt{0+1}+(0+1)$

$\Rightarrow 1+1$

$\Rightarrow 2$

Hence, this is the answer.