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

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(A) Using the binomial expansion for $(1+x)^n$ where $x 	o 0$:$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ...$Substituting this into the limit:$\matho

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$n$

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(A) Using the binomial expansion for $(1+x)^n$ where $x 	o 0$:$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ...$Substituting this into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{(1 + nx + \frac{n(n-1)}{2}x^2 + ...) - 1}{x}$$= \mathop {\lim }\limits_{x 	o 0} \frac{nx + \frac{n(n-1)}{2}x^2 + ...}{x}$$= \mathop {\lim }\limits_{x 	o 0} (n + \frac{n(n-1)}{2}x + ...)$$= n$Alternatively,applying $L-Hospital's$ rule by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}((1+x)^n - 1)}{\frac{d}{dx}(x)} = \mathop {\lim }\limits_{x 	o 0} \frac{n(1+x)^{n-1}}{1} = n(1+0)^{n-1} = n$.

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

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