mathop {lim }limits_{x to 1} frac{{1 - {x^{ - | vedclass

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Question

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}}$.છેદને અવયવ પાડતા: $1 - {x^{ - 2/3}} = (1 - {x^{ - 1/3}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}}$.છેદને અવયવ પાડતા: $1 - {x^{ - 2/3}} = (1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})$.લક્ષમાં કિંમત મુકતા:$\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{(1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})}}$.સામાન્ય પદ $(1 - {x^{ - 1/3}})$ ઉડાડતા:$\mathop {\lim }\limits_{x 	o 1} \frac{1}{{1 + {x^{ - 1/3}}}} = \frac{1}{{1 + 1^{ - 1/3}}} = \frac{1}{{1 + 1}} = \frac{1}{2}$.વૈકલ્પિક રીતે,$L$-Hospital ના નિયમનો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(1 - x^{-1/3})}{\frac{d}{dx}(1 - x^{-2/3})} = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{3}x^{-4/3}}{\frac{2}{3}x^{-5/3}} = \mathop {\lim }\limits_{x 	o 1} \frac{1}{2}x^{1/3} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

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