If the function f(x) = left(frac{5x-8}{8-3x}r | vedclass

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(A) Given that $f(x)$ is continuous at $x = 2$,we have $f(2) = \lim_{x \rightarrow 2} f(x)$.$k = \lim_{x \rightarrow 2} \left(\frac{5x-8}{8-3x}\right)

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$e^6$

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(A) Given that $f(x)$ is continuous at $x = 2$,we have $f(2) = \lim_{x \rightarrow 2} f(x)$.$k = \lim_{x \rightarrow 2} \left(\frac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}$.Let $x - 2 = h$,so $x = 2 + h$. As $x \rightarrow 2$,$h \rightarrow 0$.$k = \lim_{h \rightarrow 0} \left(\frac{5(2+h)-8}{8-3(2+h)}\right)^{\frac{3}{2(2+h)-4}} = \lim_{h \rightarrow 0} \left(\frac{10+5h-8}{8-6-3h}\right)^{\frac{3}{2h}} = \lim_{h \rightarrow 0} \left(\frac{2+5h}{2-3h}\right)^{\frac{3}{2h}}$.$k = \lim_{h \rightarrow 0} \left(\frac{2(1 + \frac{5}{2}h)}{2(1 - \frac{3}{2}h)}\right)^{\frac{3}{2h}} = \lim_{h \rightarrow 0} \frac{(1 + \frac{5}{2}h)^{\frac{3}{2h}}}{(1 - \frac{3}{2}h)^{\frac{3}{2h}}}$.Using the formula $\lim_{u \rightarrow 0} (1+u)^{\frac{1}{u}} = e$,we have:Numerator: $\lim_{h \rightarrow 0} [(1 + \frac{5}{2}h)^{\frac{1}{\frac{5}{2}h}}]^{\frac{5}{2} \cdot \frac{3}{2h} \cdot h} = e^{\frac{15}{4}}$.Denominator: $\lim_{h \rightarrow 0} [(1 - \frac{3}{2}h)^{\frac{1}{-\frac{3}{2}h}}]^{-\frac{3}{2} \cdot \frac{3}{2h} \cdot h} = e^{-\frac{9}{4}}$.Thus,$k = \frac{e^{15/4}}{e^{-9/4}} = e^{\frac{15}{4} + \frac{9}{4}} = e^{\frac{24}{4}} = e^6$.

If the function $f(x) = \left(\frac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}$ for $x \neq 2$ and $f(2) = k$ is continuous at $x = 2$,then $k =$

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