If f(x) = begin{cases} frac{8^x - 4^x - 2^x + | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0)$.First,calculate the right-hand limit: $\lim_{x 	o 0

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$1000$

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(A) For $f(x)$ to be continuous at $x = 0$,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0)$.First,calculate the right-hand limit: $\lim_{x 	o 0^+} \frac{8^x - 4^x - 2^x + 1}{x^2} = \lim_{x 	o 0^+} \frac{(4^x - 1)(2^x - 1)}{x^2} = \lim_{x 	o 0^+} \left( \frac{4^x - 1}{x} ight) \left( \frac{2^x - 1}{x} ight) = \log 4 \cdot \log 2$.Next,calculate the left-hand limit: $\lim_{x 	o 0^-} (e^x \sin x + kx + \lambda \log 4) = e^0 \sin 0 + k(0) + \lambda \log 4 = \lambda \log 4$.Equating the two: $\lambda \log 4 = \log 4 \cdot \log 2 \implies \lambda = \log 2$.Then $e^\lambda = e^{\log 2} = 2$.Finally,$500 e^\lambda = 500 	imes 2 = 1000$.

If $f(x) = \begin{cases} \frac{8^x - 4^x - 2^x + 1}{x^2} & , \text{if } x > 0 \\ e^x \sin x + kx + \lambda \log 4 & , \text{if } x \le 0 \end{cases}$ is continuous at $x = 0$,then the value of $500 e^\lambda$ is

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