Let the function f(x) = begin{cases} frac{log | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must be equal to $f(0)$.Given $f(0) = 10$,we have:$\lim_{x

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$-5$

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must be equal to $f(0)$.Given $f(0) = 10$,we have:$\lim_{x 	o 0} \frac{\ln(1+5x) - \ln(1+\alpha x)}{x} = 10$Using the standard limit formula $\lim_{x 	o 0} \frac{\ln(1+kx)}{x} = k$,we can write:$\lim_{x 	o 0} \left( \frac{\ln(1+5x)}{x} - \frac{\ln(1+\alpha x)}{x} ight) = 10$Applying the limit:$5 - \alpha = 10$Solving for $\alpha$:$\alpha = 5 - 10 = -5$Thus,the value of $\alpha$ is $-5$.

Let the function $f(x) = \begin{cases} \frac{\log_{e}(1+5x) - \log_{e}(1+\alpha x)}{x} & \text{if } x \neq 0 \\ 10 & \text{if } x = 0 \end{cases}$ be continuous at $x = 0$. The value of $\alpha$ is equal to:

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