The true statement for mathop {lim }limits_{x | vedclass

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(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$.Rationalizing the numerator

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$.Rationalizing the numerator and denominator:$L = \mathop {\lim }\limits_{x 	o 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{(\sqrt{2+3x} - \sqrt{2-3x})(\sqrt{2+3x} + \sqrt{2-3x})} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{(1+x) - (1-x)}{(2+3x) - (2-3x)} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{2x}{6x} 	imes \frac{\sqrt{2+3x} + \sqrt{2-3x}}{\sqrt{1+x} + \sqrt{1-x}}$$L = \frac{1}{3} 	imes \frac{\sqrt{2} + \sqrt{2}}{1 + 1} = \frac{1}{3} 	imes \frac{2\sqrt{2}}{2} = \frac{\sqrt{2}}{3}$.Since $\sqrt{2} \approx 1.414$,then $\frac{\sqrt{2}}{3} \approx \frac{1.414}{3} \approx 0.471$.Thus,$0 Therefore,the correct option is $B$.

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

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