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

The correct option is C $2$
$\underset{x\to \infty }{lim}[x\sqrt{x^2+4}-\sqrt{x^4+16}]$
Rationalising the terms inside the bracket
$=\underset{x\to \infty }{lim}[x\sqrt{x^2+4}-\sqrt{x^4+16}\times \frac{x\sqrt{x^2+4}+\sqrt{x^4+16}}{x\sqrt{x^2+4}+\sqrt{x^4+16}}]$

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

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

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

$=\frac{4}{1+1}$
$=2$