mathop {lim }limits_{x to 3} left{ {frac{{x - | vedclass

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Question

(A) લક્ષ $\mathop {\lim }\limits_{x 	o 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}$ ની ગણતરી કરવા માટે,છેદનું સંમેયીકરણ કરતા:$\mathop {\lim }\limits

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(A) લક્ષ $\mathop {\lim }\limits_{x 	o 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}$ ની ગણતરી કરવા માટે,છેદનું સંમેયીકરણ કરતા:$\mathop {\lim }\limits_{x 	o 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2} - \sqrt{4 - x})(\sqrt{x - 2} + \sqrt{4 - x})}$$= \mathop {\lim }\limits_{x 	o 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(x - 2) - (4 - x)}$$= \mathop {\lim }\limits_{x 	o 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2x - 6}$$= \mathop {\lim }\limits_{x 	o 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2(x - 3)}$$= \mathop {\lim }\limits_{x 	o 3} \frac{\sqrt{x - 2} + \sqrt{4 - x}}{2}$$= \frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2} = \frac{1 + 1}{2} = 1$.વૈકલ્પિક રીતે,$L$-Hospital નો નિયમ વાપરતા:$\mathop {\lim }\limits_{x 	o 3} \frac{\frac{d}{dx}(x - 3)}{\frac{d}{dx}(\sqrt{x - 2} - \sqrt{4 - x})} = \mathop {\lim }\limits_{x 	o 3} \frac{1}{\frac{1}{2\sqrt{x - 2}} - \frac{-1}{2\sqrt{4 - x}}} = \frac{1}{\frac{1}{2} + \frac{1}{2}} = 1$.

$\mathop {\lim }\limits_{x \to 3} \left\{ {\frac{{x - 3}}{{\sqrt {x - 2} - \sqrt {4 - x} }}} \right\} = $

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