If mathop {lim }limits_{x to 2} frac{{{x^n} - | vedclass

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(B) We know the standard limit formula: $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n \cdot a^{n-1}$.Given the expression: $\m

Vedclass · Accepted Answer

$5$

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(B) We know the standard limit formula: $\mathop {\lim }\limits_{x 	o a} \frac{{{x^n} - {a^n}}}{{x - a}} = n \cdot a^{n-1}$.Given the expression: $\mathop {\lim }\limits_{x 	o 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$.Applying the formula with $a = 2$,we get: $n \cdot 2^{n-1} = 80$.We can write $80$ as $5 \cdot 16 = 5 \cdot 2^4$.Comparing $n \cdot 2^{n-1} = 5 \cdot 2^4$,we find $n = 5$.

If $\mathop {\lim }\limits_{x \to 2} \frac{{{x^n} - {2^n}}}{{x - 2}} = 80$,where $n$ is a positive integer,then $n = $

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