mathop {lim }limits_{x to 0} frac{{{{(1 + x)} | vedclass

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(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x}$,we can use the rationalization method o

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(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x}$,we can use the rationalization method or $L$-Hospital's rule.Method $1$: RationalizationMultiply the numerator and denominator by the conjugate $\frac{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}}$:$\mathop {\lim }\limits_{x 	o 0} \frac{{(1 + x) - (1 - x)}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2x}}{{x({{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}})}} = \mathop {\lim }\limits_{x 	o 0} \frac{2}{{{{(1 + x)}^{1/2}} + {{(1 - x)}^{1/2}}}} = \frac{2}{{1 + 1}} = 1$.Method $2$: $L$-Hospital's ruleSince the limit is in the form $\frac{0}{0}$,differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}})}}{{\frac{d}{{dx}}(x)}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{2}{{(1 + x)}^{ - 1/2}} - ( - \frac{1}{2}{{(1 - x)}^{ - 1/2}})}}{1} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }} ight) = \frac{1}{2} + \frac{1}{2} = 1$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x} = $

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