mathop {lim }limits_{x to 0} frac{{sqrt {3 + | vedclass

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(D) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x}$,we rationalize the numerator:Multiply the nume

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$\frac{1}{\sqrt{3}}$

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(D) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x}$,we rationalize the numerator:Multiply the numerator and denominator by $(\sqrt{3+x} + \sqrt{3-x})$:$= \mathop {\lim }\limits_{x 	o 0} \frac{(\sqrt{3+x} - \sqrt{3-x})(\sqrt{3+x} + \sqrt{3-x})}{x(\sqrt{3+x} + \sqrt{3-x})}$$= \mathop {\lim }\limits_{x 	o 0} \frac{(3+x) - (3-x)}{x(\sqrt{3+x} + \sqrt{3-x})}$$= \mathop {\lim }\limits_{x 	o 0} \frac{2x}{x(\sqrt{3+x} + \sqrt{3-x})}$$= \mathop {\lim }\limits_{x 	o 0} \frac{2}{\sqrt{3+x} + \sqrt{3-x}}$$= \frac{2}{\sqrt{3+0} + \sqrt{3-0}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {3 + x} - \sqrt {3 - x} }}{x} = $

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