mathop {lim}limits_{x to 1} left[ {left[ {fra | vedclass

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Question

(B) Let $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = \frac{4}{{{x^2} - \frac{1}{x}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x}}{{{x^3} - 1}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}} = \frac{{4x + 1 - 3x + {x^2}}}{{{x^3} - 1}} = \frac{{{x^2} + x + 1}}{{(x - 1)({x^2} + x + 1)}} = \frac{1}{{x - 1}}$.Then,the first term is $[f(x)]^{-1} = x - 1$.The second term is $\frac{{3({x^4} - 1)}}{{{x^3} - \frac{1}{x}}} = \frac{{3x({x^4} - 1)}}{{{x^4} - 1}} = 3x$.Thus,the expression becomes $\mathop {\lim}\limits_{x 	o 1} [(x - 1) + 3x] = \mathop {\lim}\limits_{x 	o 1} (4x - 1) = 4(1) - 1 = 3$.

$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

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