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(D) Since $f(x)$ is continuous at $x = 0$, we have $\lim_{x 	o 0} f(x) = f(0)$.$\lim_{x 	o 0} \frac{8^x - 2^x}{k^x - 1} = 2$.Divide numerator and de

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$2$

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(D) Since $f(x)$ is continuous at $x = 0$, we have $\lim_{x 	o 0} f(x) = f(0)$.$\lim_{x 	o 0} \frac{8^x - 2^x}{k^x - 1} = 2$.Divide numerator and denominator by $x$: $\lim_{x 	o 0} \frac{\frac{8^x - 1}{x} - \frac{2^x - 1}{x}}{\frac{k^x - 1}{x}} = 2$.Using the standard limit $\lim_{x 	o 0} \frac{a^x - 1}{x} = \ln a$, we get $\frac{\ln 8 - \ln 2}{\ln k} = 2$.$\frac{\ln(8/2)}{\ln k} = 2 \implies \frac{\ln 4}{\ln k} = 2$.$\ln 4 = 2 \ln k \implies \ln 4 = \ln(k^2)$.$k^2 = 4 \implies k = 2$ (since $k^x$ is defined for $k > 0$ and $k 
eq 1$ for the limit to exist).

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...

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