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Let $f(x)=\left\{\begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k, & x=2\end{array}\right.$
The value of $k$ for which $f$ is continuous at $x=2$ is
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The value of $k$ for which $f$ is continuous at $x=2$ is
Solution:
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Verified Answer
The correct answer is:
$e^{-1}$
$e^{-1}$
Since $f(x)$ is continuous at $x=2$.
$$
\begin{aligned}
&\therefore \lim _{x \rightarrow 2} f(x)=f(2) \\
&\Rightarrow \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=k \quad\left(1^{\infty} \text { form }\right) \\
&\therefore e^{l=k}
\end{aligned}
$$
where
$$
\begin{aligned}
&l=\lim _{x \rightarrow 2}(x-1-1) \times \frac{1}{2-x}=\lim _{x \rightarrow 2} \frac{x-2}{2-x} \\
&=\lim _{x \rightarrow 2}\left(\frac{x-2}{x-2}\right) \\
&\Rightarrow k=e^{-1}
\end{aligned}
$$
$$
\begin{aligned}
&\therefore \lim _{x \rightarrow 2} f(x)=f(2) \\
&\Rightarrow \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=k \quad\left(1^{\infty} \text { form }\right) \\
&\therefore e^{l=k}
\end{aligned}
$$
where
$$
\begin{aligned}
&l=\lim _{x \rightarrow 2}(x-1-1) \times \frac{1}{2-x}=\lim _{x \rightarrow 2} \frac{x-2}{2-x} \\
&=\lim _{x \rightarrow 2}\left(\frac{x-2}{x-2}\right) \\
&\Rightarrow k=e^{-1}
\end{aligned}
$$
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