Let f(x) = begin{cases} frac{(1 + tan x)^{fra | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x 	o 0} f(x)$.$k = \lim_{x 	o 0} \frac{(1 + 	an x)^{\frac{1}{x}} - e}{x} = \lim_

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$-\frac{e}{2}$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $k = \lim_{x 	o 0} f(x)$.$k = \lim_{x 	o 0} \frac{(1 + 	an x)^{\frac{1}{x}} - e}{x} = \lim_{x 	o 0} \frac{e^{\frac{1}{x} \ln(1 + 	an x)} - e}{x}$.Factoring out $e$,we get $k = e \lim_{x 	o 0} \frac{e^{\frac{1}{x} \ln(1 + 	an x) - 1} - 1}{x}$.Using the limit $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$,we multiply and divide by $(\frac{1}{x} \ln(1 + 	an x) - 1)$:$k = e \lim_{x 	o 0} \left( \frac{e^{\frac{1}{x} \ln(1 + 	an x) - 1} - 1}{\frac{1}{x} \ln(1 + 	an x) - 1} ight) \cdot \left( \frac{\ln(1 + 	an x) - x}{x^2} ight)$.Using the expansion $\ln(1 + t) = t - \frac{t^2}{2} + \dots$ and $	an x = x + \frac{x^3}{3} + \dots$,we have $\ln(1 + 	an x) = (x + \frac{x^3}{3}) - \frac{(x + \dots)^2}{2} = x - \frac{x^2}{2} + \dots$.Thus,$\lim_{x 	o 0} \frac{\ln(1 + 	an x) - x}{x^2} = \lim_{x 	o 0} \frac{x - \frac{x^2}{2} - x}{x^2} = -\frac{1}{2}$.Therefore,$k = e \cdot 1 \cdot (-\frac{1}{2}) = -\frac{e}{2}$.

Let $f(x) = \begin{cases} \frac{(1 + \tan x)^{\frac{1}{x}} - e}{x} & x \neq 0 \\ k & x = 0 \end{cases}$ be continuous at $x = 0$,then the value of $k$ is:

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