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) $\mathop {\lim }\limits_{x \to \infty } \,\left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{x + \sqrt {x + \sqrt x } - x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}$
(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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