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

Answer

Correct option: C.
$2$
c
(c) On rationalization $\mathop {\lim }\limits_{x \to \infty } \,\frac{{4x}}{{(\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3} }}$

$ = \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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