MCQ
The value of $\mathop {\lim }\limits_{x \to \infty } \sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1} $ is
  • $\frac{1}{2}$
  • B
    $1$
  • C
    $2$
  • D
    None of these

Answer

Correct option: A.
$\frac{1}{2}$
a
(a) $\mathop {\lim }\limits_{x \to \infty } \,\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1} $

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{{ax}}{{\,\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1} }}$

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{a}{{\,\sqrt {{a^2} + \frac{a}{x} + \frac{1}{{{x^2}}}} + \sqrt {{a^2} + \frac{1}{{{x^2}}}} }} = \frac{a}{{2a}} = \frac{1}{2}$.

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