For the function f(x) = frac{log_e(1 + x) - l | vedclass

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(D) For a function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$

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$2$

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(D) For a function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Given $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$.As $x 	o 0$,the expression takes the indeterminate form $\frac{0}{0}$.Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{\frac{d}{dx}(\log_e(1 + x) - \log_e(1 - x))}{\frac{d}{dx}(x)}$$= \lim_{x 	o 0} \frac{\frac{1}{1 + x} - (\frac{-1}{1 - x})}{1}$$= \lim_{x 	o 0} (\frac{1}{1 + x} + \frac{1}{1 - x})$$= \frac{1}{1 + 0} + \frac{1}{1 - 0} = 1 + 1 = 2$.Thus,for the function to be continuous at $x = 0$,$f(0)$ must be $2$.

For the function $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$ to be continuous at $x = 0$,the value of $f(0)$ should be

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