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

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Question

(A) To evaluate the limit $\mathop {\lim }\limits_{h 	o 0} \frac{{\sqrt {x + h} - \sqrt x }}{h}$,we rationalize the numerator:$\mathop {\lim }\limits

Vedclass · Accepted Answer

$\frac{1}{{2\sqrt x }}$

Vedclass · Answer

(A) To evaluate the limit $\mathop {\lim }\limits_{h 	o 0} \frac{{\sqrt {x + h} - \sqrt x }}{h}$,we rationalize the numerator:$\mathop {\lim }\limits_{h 	o 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} 	imes \frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }} = \mathop {\lim }\limits_{h 	o 0} \frac{{(x + h) - x}}{{h(\sqrt {x + h} + \sqrt x )}}$$= \mathop {\lim }\limits_{h 	o 0} \frac{h}{{h(\sqrt {x + h} + \sqrt x )}} = \mathop {\lim }\limits_{h 	o 0} \frac{1}{{\sqrt {x + h} + \sqrt x }}$$= \frac{1}{{\sqrt {x + 0} + \sqrt x }} = \frac{1}{{2\sqrt x }}$Alternatively,using $L'	ext{Hospital's rule}$ by differentiating with respect to $h$:$\mathop {\lim }\limits_{h 	o 0} \frac{{\frac{d}{{dh}}(\sqrt {x + h} - \sqrt x )}}{{\frac{d}{{dh}}(h)}} = \mathop {\lim }\limits_{h 	o 0} \frac{{\frac{1}{{2\sqrt {x + h} }}}}{1} = \frac{1}{{2\sqrt x }}$.

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

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