Let the function f(x) be defined as: f(x) = b | vedclass

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(B) For $f(x)$ to be continuous at $x = 0$, we must have $k = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} [	an(\frac{\pi}{4} + x)]^{\frac{1}{x}}$ is of the

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$e^2$

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(B) For $f(x)$ to be continuous at $x = 0$, we must have $k = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} [	an(\frac{\pi}{4} + x)]^{\frac{1}{x}}$ is of the form $1^\infty$.Using the formula $\lim_{x 	o 0} [g(x)]^{h(x)} = e^{\lim_{x 	o 0} h(x)[g(x) - 1]}$, we get:$k = e^{\lim_{x 	o 0} \frac{1}{x} [	an(\frac{\pi}{4} + x) - 1]}$.Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, we have $	an(\frac{\pi}{4} + x) = \frac{1 + 	an x}{1 - 	an x}$.So, $k = e^{\lim_{x 	o 0} \frac{1}{x} [\frac{1 + 	an x}{1 - 	an x} - 1]} = e^{\lim_{x 	o 0} \frac{1}{x} [\frac{1 + 	an x - 1 + 	an x}{1 - 	an x}]} = e^{\lim_{x 	o 0} \frac{2 	an x}{x(1 - 	an x)}}$.Since $\lim_{x 	o 0} \frac{	an x}{x} = 1$, we get $k = e^{\frac{2(1)}{1 - 0}} = e^2$.

Let the function $f(x)$ be defined as: $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ k, & x = 0 \end{cases}$. If $f(x)$ is continuous at $x = 0$, then the value of $k$ is...

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