If mathop {lim }limits_{x to 5} frac{{{x^k} - | vedclass

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(B) We know the standard limit formula: $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$.Applying this to the given l

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$4$

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(B) We know the standard limit formula: $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$.Applying this to the given limit: $\mathop {\lim }\limits_{x 	o 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = k{(5)^{k - 1}}$.Given that the limit equals $500$,we have: $k{(5)^{k - 1}} = 500$.We can rewrite $500$ as $4 	imes 125 = 4 	imes 5^3 = 4 	imes 5^{4-1}$.Comparing $k{(5)^{k - 1}}$ with $4{(5)^{4 - 1}}$,we get $k = 4$.

If $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = 500$,then the positive integral value of $k$ is

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