MCQ
$\mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$ is equal to
  • A
    $0$
  • $\frac{1}{2}$
  • C
    $log 2$
  • D
    ${e^4}$

Answer

Correct option: B.
$\frac{1}{2}$
b
(b) $\mathop {\lim }\limits_{x \to \infty } \,\left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{x + \sqrt {x + \sqrt x } - x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {x + \sqrt x } }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \,\frac{{\sqrt {1 + {x^{ - 1/2}}} }}{{\sqrt {1 + \sqrt {{x^{ - 1}} + {x^{ - 3/2}}} } + 1}} = \frac{1}{2}$.

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