Question

The value of $\underset{x\to \infty }{\lim }[x\sqrt{x^2+4}-\sqrt{x^4+16}]$ is

• A. 4
• B. 8
• C. 2
• D. 16

Answer 1

Answer: (C)

We have,
$\underset{x\to \infty }{\lim }\left[x\sqrt{x^2+4}-\sqrt{x^4+16}\right]$
$=\underset{x\to \infty }{\lim }\left[x\sqrt{x^2+4}-x^4+16\right]$
$\left(\frac{x\sqrt{x^2+4}+\sqrt{x^4+16}}{x\sqrt{x^2+4}+\sqrt{x^4+16}}\right)$
$=\underset{x\to \infty }{\lim }\frac{x^2\left(x^2+4\right)-\left(x^4+16\right)}{x\sqrt{x^2+4}+\sqrt{x^4+16}}$
$=\underset{x\to \infty }{\lim }\frac{4x^2-16}{\left(\sqrt{1+\frac{4}{x^2}+\sqrt{1+\frac{16}{x^4}}}\right)}$
$=\underset{x\to \infty }{\lim }\frac{4-\frac{16}{x^2}}{\sqrt{1+\frac{4}{x^2}+\sqrt{1+\frac{16}{x^4}}}}=\frac{4}{2}=2$