The value of f(0) so that the function f(x) = | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{2^x - 2^{-x}}{x}$This is a $0/

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$\log 4$

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{2^x - 2^{-x}}{x}$This is a $0/0$ form,so we apply $L$'$H$ôpital's rule:$\lim_{x 	o 0} \frac{\frac{d}{dx}(2^x - 2^{-x})}{\frac{d}{dx}(x)} = \lim_{x 	o 0} \frac{2^x \ln 2 - 2^{-x} \ln 2(-1)}{1}$$= \lim_{x 	o 0} (2^x \ln 2 + 2^{-x} \ln 2)$$= (2^0 \ln 2 + 2^0 \ln 2) = \ln 2 + \ln 2 = 2 \ln 2$Using the property $n \ln a = \ln(a^n)$,we get $2 \ln 2 = \ln(2^2) = \ln 4$.

The value of $f(0)$ so that the function $f(x) = \frac{2^x - 2^{-x}}{x}$ for $x \neq 0$ is continuous at $x = 0$ is:

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