If f(x) = begin{cases} k, & text{for } x = 1 | vedclass

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(D) Since $f(x)$ is continuous at $x = 1$, we have $k = \lim_{x \rightarrow 1} f(x)$.$k = \lim_{x \rightarrow 1} \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$

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$\frac{1}{2}$

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(D) Since $f(x)$ is continuous at $x = 1$, we have $k = \lim_{x \rightarrow 1} f(x)$.$k = \lim_{x \rightarrow 1} \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$.Using $L$'$H$ôpital's rule, we differentiate the numerator and denominator with respect to $x$:Numerator: $\frac{d}{dx}[(9x-1)(\sqrt{x}-1)] = 9(\sqrt{x}-1) + (9x-1)(\frac{1}{2\sqrt{x}})$.Denominator: $\frac{d}{dx}[3x^2+2x-5] = 6x+2$.Evaluating the limit as $x \rightarrow 1$:$k = \frac{9(1-1) + (9-1)(\frac{1}{2})}{6(1)+2} = \frac{0 + 8(\frac{1}{2})}{8} = \frac{4}{8} = \frac{1}{2}$.

If $f(x) = \begin{cases} k, & \text{for } x = 1 \\ \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}, & \text{for } x \neq 1 \end{cases}$ is continuous on $[0, \infty)$, then $k =$

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