Let k be a non-zero real number. If f(x) = be | vedclass

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(C) Since $f(x)$ is a continuous function at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{(e^x - 1)^2

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$3$

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(C) Since $f(x)$ is a continuous function at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{(e^x - 1)^2}{\sin(\frac{x}{k}) \log(1 + \frac{x}{4})}$We know that $\lim_{x 	o 0} \frac{e^x - 1}{x} = 1$,$\lim_{x 	o 0} \frac{\sin(\frac{x}{k})}{\frac{x}{k}} = 1$,and $\lim_{x 	o 0} \frac{\log(1 + \frac{x}{4})}{\frac{x}{4}} = 1$.Rewriting the expression:$\lim_{x 	o 0} \frac{x^2 (\frac{e^x - 1}{x})^2}{(\frac{x}{k}) \frac{\sin(\frac{x}{k})}{\frac{x}{k}} \cdot (\frac{x}{4}) \frac{\log(1 + \frac{x}{4})}{\frac{x}{4}}}$$= \lim_{x 	o 0} \frac{x^2 (\frac{e^x - 1}{x})^2}{\frac{x^2}{4k} \cdot \frac{\sin(\frac{x}{k})}{\frac{x}{k}} \cdot \frac{\log(1 + \frac{x}{4})}{\frac{x}{4}}}$$= 4k \cdot \frac{1^2}{1 \cdot 1} = 4k$.Given $f(0) = 12$,we have $4k = 12$,which implies $k = 3$.

Let $k$ be a non-zero real number. If $f(x) = \begin{cases} \frac{(e^x - 1)^2}{\sin(\frac{x}{k}) \log(1 + \frac{x}{4})} & x \neq 0 \\ 12 & x = 0 \end{cases}$ is a continuous function,then the value of $k$ is:

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