Find k if the following function is continuous at the point indicated against them: . rlrl f(x) & = (5 x-8 8-3 x )^ 3 2 x-4 , & & for x 2 & =k_1 & & for x=2 at x=2

f(x) is continuous at x=2 & f(2)= _ x 2 f(x) & k= _ x 2 (5 x-8 8-3 x )^ 3 2 x-4 Put x-2=h x=2+h As x 2, ~h 0 & k= _ h 0 [5(2+h)-8 8-3(2+h) ]^ 3 2 h & = _ h 0 (10+5 h-8 8-6-3 h )^ 3 2 h & = _ h 0 (2+5 h 2-3 h )^ 3 2 h & = _ h 0 [ 2 (1+5 h 2 ) 2 (1-3 h 2 ) ]^ 3 2 h & = _ h 0 (1+5 h 2 )^ 3 2 h (1-3 h 2 )^ 3 2 h & = _ h 0 [ (1+5 h 2 )^ 2 3 h ]^ 5 2 3 2 _ h 0 [ (1-3 h 2 )^ -2 3 h ]^ -3 2 3 2 & = e^ 15 4 e^ -9 4 [ h 0, 5 h 2 0, -3 h 2 0 ] & =e^ 24 4 and _ x 0 (1+x)^ 1 2 =e & =e^6

Find k if the following function is continuous at the point indic…

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Question

Find $k$ if the following function is continuous at the point indicated against them:
$
\left.\begin{array}{rlrl}
f(x) & =\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}, & & \text { for } x \neq 2 \\
& =k_1 & & \text { for } x=2
\end{array}\right\} \text { at } x=2
$

✓

Answer

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