If the function f(x) = begin{cases} frac{(e^{ | vedclass

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(D) Given that the function $f(x)$ is continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Since $f(0) = 16$,we evaluate the limit:$\lim_{x

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$\pm 8$

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(D) Given that the function $f(x)$ is continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Since $f(0) = 16$,we evaluate the limit:$\lim_{x 	o 0} \frac{(e^{kx} - 1) 	an kx}{4x^2} = 16$.We can rewrite the expression as:$\lim_{x 	o 0} \left( \frac{e^{kx} - 1}{kx} ight) \left( \frac{	an kx}{kx} ight) \left( \frac{k^2 x^2}{4x^2} ight) = 16$.Using the standard limits $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$ and $\lim_{u 	o 0} \frac{	an u}{u} = 1$,we get:$1 	imes 1 	imes \frac{k^2}{4} = 16$.$\frac{k^2}{4} = 16$.$k^2 = 64$.$k = \pm 8$.

If the function $f(x) = \begin{cases} \frac{(e^{kx} - 1) \tan kx}{4x^2}, & x \neq 0 \\ 16, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = . . . . . .$.

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