Answer
$$\lim\limits_{x\to\infty}\frac{x^2}{\sqrt{x^4+1}}=1$$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{x^2}{\sqrt{x^4+1}}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$
Divide both numerator and denominator by $x^2$, we have:
- In the numerator:
$X=\frac{x^2}{x^2}=1$
- In the denominator:
$Y=\frac{\sqrt{x^4+1}}{x^2}=\frac{\sqrt{x^4+1}}{\sqrt{x^4}}=\sqrt{\frac{x^4+1}{x^4}}=\sqrt{1+\frac{1}{x^4}}$
Therefore, $$A=\lim\limits_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^4}}}$$$$A=\frac{1}{\lim\limits_{x\to\infty}\sqrt{1+\frac{1}{x^4}}}$$$$A=\frac{1}{\sqrt{1+\lim\limits_{x\to\infty}(\frac{1}{x^4})}}$$$$A=\frac{1}{\sqrt{1+0}}$$$$A=1$$
