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

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Question

(B) We are given the limit: $\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}}$.Notice that the denominator can be factor

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) We are given the limit: $\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}}$.Notice that the denominator can be factored as a difference of squares: $1 - {x^{ - 2/3}} = (1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})$.Substituting this into the limit expression:$\mathop {\lim }\limits_{x 	o 1} \frac{{1 - {x^{ - 1/3}}}}{{(1 - {x^{ - 1/3}})(1 + {x^{ - 1/3}})}}$.Canceling the common term $(1 - {x^{ - 1/3}})$ (since $x 	o 1$,$x 
eq 1$):$\mathop {\lim }\limits_{x 	o 1} \frac{1}{{1 + {x^{ - 1/3}}}} = \frac{1}{{1 + 1^{ - 1/3}}} = \frac{1}{{1 + 1}} = \frac{1}{2}$.Alternatively,applying $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(1 - x^{-1/3})}{\frac{d}{dx}(1 - x^{-2/3})} = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{3}x^{-4/3}}{\frac{2}{3}x^{-5/3}} = \mathop {\lim }\limits_{x 	o 1} \frac{1}{2}x^{1/3} = \frac{1}{2}$.

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

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Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
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