If mathop {lim }limits_{x to 1} frac{{{x^4} - | vedclass

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(B) First,evaluate the left-hand side limit:$\mathop {\lim }\limits_{x 	o 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x 	o 1} \frac{(x-

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$\frac{8}{3}$

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(B) First,evaluate the left-hand side limit:$\mathop {\lim }\limits_{x 	o 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x 	o 1} \frac{(x-1)(x+1)(x^2+1)}{x-1} = \mathop {\lim }\limits_{x 	o 1} (x+1)(x^2+1) = (1+1)(1^2+1) = 2 	imes 2 = 4$.Next,evaluate the right-hand side limit:$\mathop {\lim }\limits_{x 	o k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \mathop {\lim }\limits_{x 	o k} \frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)} = \mathop {\lim }\limits_{x 	o k} \frac{x^2+xk+k^2}{x+k} = \frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}$.Equating both sides:$4 = \frac{3k}{2}$ $\Rightarrow 3k = 8$ $\Rightarrow k = \frac{8}{3}$.

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ is

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