If f(x) = begin{cases} frac{81^{x}-9^{x}}{k^{ | vedclass

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(A) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.$\lim _{x \rightarrow 0} f(x) = f(0

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$3$

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(A) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.$\lim _{x \rightarrow 0} f(x) = f(0) = 2$.$\lim _{x \rightarrow 0} \frac{81^{x}-9^{x}}{k^{x}-1} = 2$.Factor out $9^{x}$ from the numerator: $\lim _{x \rightarrow 0} \frac{9^{x}(9^{x}-1)}{k^{x}-1} = 2$.Divide the numerator and denominator by $x$: $\lim _{x \rightarrow 0} \frac{9^{x} \cdot \frac{9^{x}-1}{x}}{\frac{k^{x}-1}{x}} = 2$.Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} = \ln a$,we get:$\frac{9^{0} \cdot \ln 9}{\ln k} = 2$.Since $9^{0} = 1$,we have $\frac{\ln 9}{\ln k} = 2$.$\ln 9 = 2 \ln k \Rightarrow \ln 9 = \ln k^{2}$.$k^{2} = 9 \Rightarrow k = 3$ (since $k$ must be positive for the base of an exponential function).Thus,the value of $k$ is $3$.

If $f(x) = \begin{cases} \frac{81^{x}-9^{x}}{k^{x}-1} & x \neq 0 \\ 2 & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

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