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

Given, \(\mathrm{f}(x)\) is continuous at \(x=2\)
\(\begin{array}{ll}
\therefore \quad & f(2)=\lim _{x \rightarrow 2}(x) \\
& \mathrm{k}=\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}} \\
& \text { Put } x-2=\mathrm{h} \\
& \Rightarrow x=2+\mathrm{h} \\
& \text { As } x \rightarrow 2, \mathrm{~h} \rightarrow 0 \\
\therefore \quad & \mathrm{k}=\lim _{\mathrm{h} \rightarrow 0}\left[\frac{5(2+\mathrm{h})-8}{8-3(2+\mathrm{h})}\right]^{\frac{3}{2(2+\mathrm{h})-4}} \\
& \mathrm{k}=\lim _{\mathrm{h} \rightarrow 0}\left[\frac{10+5 \mathrm{~h}-8}{8-6-3 \mathrm{~h}}\right]^{\frac{3}{2 \mathrm{~h}}} \\
& \mathrm{k}=\lim _{\mathrm{h} \rightarrow 0}\left[\frac{2+5 \mathrm{~h}}{2-3 \mathrm{~h}}\right]^{\frac{3}{2 \mathrm{~h}}}
\end{array}\)
\(\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\left(1+\frac{5}{2} h\right)^{\frac{3}{2 h}}}{\left(1-\frac{3 h}{2}\right)^{\frac{3}{2 h}}} \\ & =\frac{\lim _{h \rightarrow 0}\left[\left(1+\frac{5}{2} h\right)^{\frac{2}{5 h}}\right]^{\frac{5}{2} \times \frac{3}{2}}}{\lim _{h \rightarrow 0}\left[\left(1-\frac{3 h}{2}\right)^{\frac{-2}{3 h}}\right]^{\frac{-3}{2} \times \frac{3}{2}}}\end{aligned}\)
\(\begin{aligned} & =\frac{e^{\frac{15}{4}}}{e^{\frac{-9}{4}}} \\ & =e^{\frac{24}{4}} \\ & =e^6\end{aligned}\)
\(\cdots\left[\begin{array}{l}h \rightarrow 0, \frac{5 h}{2} \rightarrow 0, \frac{-3 h}{2} \rightarrow 0 \\ \text { and } \lim _{h \rightarrow 0}(1+x)^{\frac{1}{x}}=e\end{array}\right]\)