Question

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

• A. 4
• B. 8
• C. 2
• D. 16

Answer 1

Answer: (C)

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