If f(x) = left(frac{1+x}{1-x}right)^{frac{1}{ | vedclass

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(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.Evaluating the limit: $\lim_{x 	o 0} \left(\frac{1+x}{1-x}ight)^{\fr

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$e^2$

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(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.Evaluating the limit: $\lim_{x 	o 0} \left(\frac{1+x}{1-x}ight)^{\frac{1}{x}}$.This is a $1^{\infty}$ indeterminate form.Using the formula $\lim_{x 	o a} [f(x)]^{g(x)} = e^{\lim_{x 	o a} g(x)[f(x)-1]}$:$\lim_{x 	o 0} f(x) = e^{\lim_{x 	o 0} \frac{1}{x} \left( \frac{1+x}{1-x} - 1 ight)}$$= e^{\lim_{x 	o 0} \frac{1}{x} \left( \frac{1+x - (1-x)}{1-x} ight)}$$= e^{\lim_{x 	o 0} \frac{1}{x} \left( \frac{2x}{1-x} ight)}$$= e^{\lim_{x 	o 0} \frac{2}{1-x}}$$= e^{\frac{2}{1-0}} = e^2$.Therefore,$f(0) = e^2$.

If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $

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