MCQ
True statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is
- ADoes not exist
- ✓Lies between $0$ and $\frac{1}{2}$
- CLies between $\frac{1}{2}$ and $1$
- DGreater then $1$
✓
Answer
Correct option: B.
Lies between $0$ and $\frac{1}{2}$
b
(b) $\mathop {\lim }\limits_{x \to 0} \,\frac{{(1 + x) - (1 - x)}}{{(2 + 3x) - (2 - 3x)}}\,\left[ {\frac{{\sqrt {2 + 3x} + \sqrt {2 - 3x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right]$
(b) $\mathop {\lim }\limits_{x \to 0} \,\frac{{(1 + x) - (1 - x)}}{{(2 + 3x) - (2 - 3x)}}\,\left[ {\frac{{\sqrt {2 + 3x} + \sqrt {2 - 3x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right]$
$ = \frac{1}{3}\,\left[ {\frac{{2\sqrt 2 }}{2}} \right] = \frac{{\sqrt 2 }}{3},\,\,0 < \frac{{\sqrt 2 }}{3} < \frac{1}{2}.$
Aliter : Apply $ L-$ Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }} $
$= \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }}}}{{\frac{3}{{2\sqrt {2 + 3x} }} + \frac{3}{{2\sqrt {2 - 3x} }}}}$
$ = \frac{{\frac{1}{2} + \frac{1}{2}}}{{\frac{3}{{2\sqrt 2 }} + \frac{3}{{2\sqrt 2 }}}} = \frac{{2\sqrt 2 }}{6} = \frac{{\sqrt 2 }}{3}$.
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