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

Question

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

Aicharya · Accepted Answer

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

If the function $\mathrm{f}(x)=\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}$ if $x \neq 2$.…

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