If f(x) = begin{cases} frac{8^x-4^x-2^x+1}{x^ | vedclass

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

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$2000$

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(C) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0)$.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 and $f(0)$:$f(0) = e^0 \sin(0) + 0 + \lambda \log 4 = \lambda \log 4$.Equating the two limits:$\log 4 \cdot \log 2 = \lambda \log 4 \implies \lambda = \log 2$.Now,calculate $1000 e^\lambda$:$1000 e^{\log 2} = 1000 	imes 2 = 2000$.

If $f(x) = \begin{cases} \frac{8^x-4^x-2^x+1}{x^2}, & \text{if } x > 0 \\ e^x \sin x + x + \lambda \log 4, & \text{if } x \leqslant 0 \end{cases}$ is continuous at $x = 0$,then the value of $1000 e^\lambda = $

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