If f(x) = left[tan left(frac{pi}{4} + xright) | vedclass

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(C) Since $f(x)$ is continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = k$. Thus,$k = \lim_{x 	o 0} \left[	an \left(\frac{\pi}{4} + x

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$e^{2}$

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(C) Since $f(x)$ is continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = k$. Thus,$k = \lim_{x 	o 0} \left[	an \left(\frac{\pi}{4} + xight)ight]^{\frac{1}{x}}$. Using the formula $	an(\frac{\pi}{4} + x) = \frac{1 + 	an x}{1 - 	an x}$,we get: $k = \lim_{x 	o 0} \left(\frac{1 + 	an x}{1 - 	an x}ight)^{\frac{1}{x}}$. This is of the form $1^{\infty}$,so we use the limit property $\lim_{x 	o 0} (1 + u(x))^{v(x)} = e^{\lim_{x 	o 0} u(x)v(x)}$. $k = \exp \left[ \lim_{x 	o 0} \frac{1}{x} \left( \frac{1 + 	an x}{1 - 	an x} - 1 ight) ight]$. $k = \exp \left[ \lim_{x 	o 0} \frac{1}{x} \left( \frac{1 + 	an x - 1 + 	an x}{1 - 	an x} ight) ight] = \exp \left[ \lim_{x 	o 0} \frac{2 	an x}{x(1 - 	an x)} ight]$. Since $\lim_{x 	o 0} \frac{	an x}{x} = 1$,we have $k = e^{2(1)/(1-0)} = e^{2}$.

If $f(x) = \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$ for $x \neq 0$ and $f(x) = k$ for $x = 0$,is continuous at $x = 0$,then $k =$

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