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

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(A) Given that $f(x)$ is continuous at $x = 0$,we have $\lim_{x 	o 0} f(x) = f(0) = k$. Since the limit is of the form $1^\infty$,we use the formula

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$e^2$

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(A) Given that $f(x)$ is continuous at $x = 0$,we have $\lim_{x 	o 0} f(x) = f(0) = k$. Since the limit is of the form $1^\infty$,we use the formula $\lim_{x 	o a} [g(x)]^{h(x)} = e^{\lim_{x 	o a} [g(x) - 1]h(x)}$. Here,$g(x) = 	an(\frac{\pi}{4} + x)$ and $h(x) = \frac{1}{x}$. $k = e^{\lim_{x 	o 0} [	an(\frac{\pi}{4} + x) - 1] \cdot \frac{1}{x}}$. Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get $	an(\frac{\pi}{4} + x) = \frac{1 + 	an x}{1 - 	an x}$. $k = e^{\lim_{x 	o 0} [\frac{1 + 	an x}{1 - 	an x} - 1] \cdot \frac{1}{x}} = e^{\lim_{x 	o 0} [\frac{1 + 	an x - 1 + 	an x}{1 - 	an x}] \cdot \frac{1}{x}}$. $k = e^{\lim_{x 	o 0} \frac{2 	an x}{x(1 - 	an x)}} = e^{2 \cdot \lim_{x 	o 0} \frac{	an x}{x} \cdot \lim_{x 	o 0} \frac{1}{1 - 	an x}}$. Since $\lim_{x 	o 0} \frac{	an x}{x} = 1$ and $\lim_{x 	o 0} \frac{1}{1 - 	an x} = 1$,we have $k = e^{2 \cdot 1 \cdot 1} = e^2$.

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 $k = \dots$

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