MCQ
$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $
- A$0$
- B$\infty $
- ✓$2$
- D$\frac{1}{2}$
$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{4}{{\left( {\sqrt {1 + \frac{8}{x} + \frac{3}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} } \right)}} = 2$.
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