mathop {lim }limits_{x to 0} frac{{{{(1 + x)} | vedclass

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(C) Using the standard limit formula $\mathop {\lim }\limits_{u 	o a} \frac{u^n - a^n}{u - a} = n a^{n-1}$:Let $u = 1 + x$. As $x 	o 0$,$u 	o 1$.Th

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$5/3$

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(C) Using the standard limit formula $\mathop {\lim }\limits_{u 	o a} \frac{u^n - a^n}{u - a} = n a^{n-1}$:Let $u = 1 + x$. As $x 	o 0$,$u 	o 1$.The expression becomes $\mathop {\lim }\limits_{u 	o 1} \frac{u^5 - 1^5}{u^3 - 1^3}$.Dividing numerator and denominator by $(u - 1)$:$\mathop {\lim }\limits_{u 	o 1} \frac{(u^5 - 1^5)/(u - 1)}{(u^3 - 1^3)/(u - 1)} = \frac{5(1)^{5-1}}{3(1)^{3-1}} = \frac{5}{3}$.Alternatively,applying $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}((1+x)^5 - 1)}{\frac{d}{dx}((1+x)^3 - 1)} = \mathop {\lim }\limits_{x 	o 0} \frac{5(1+x)^4}{3(1+x)^2} = \frac{5(1)^4}{3(1)^2} = \frac{5}{3}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^5} - 1}}{{{{(1 + x)}^3} - 1}} = $

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