True statement for _ x 0 1 + x - 1 - x 2 + 3x - 2 - 3x is - Vidyadip

MCQ

True statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is

A
Does not exist
✓
Lies between $0$ and $\frac{1}{2}$
C
Lies between $\frac{1}{2}$ and $1$
D
Greater then $1$

✓

Answer

Correct option: B.

Lies between $0$ and $\frac{1}{2}$

b
(b) $\mathop {\lim }\limits_{x \to 0} \,\frac{{(1 + x) - (1 - x)}}{{(2 + 3x) - (2 - 3x)}}\,\left[ {\frac{{\sqrt {2 + 3x} + \sqrt {2 - 3x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right]$

$ = \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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