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If
$f(x)=\left\{\begin{array}{cl}\frac{\left.x^{2}-(a+2) x+a\right)}{x-2} & x \neq 2 \\ 2 & , x=2\end{array} \quad\right.$ is continuous at $x=2$, then the value of $a$ is
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$f(x)=\left\{\begin{array}{cl}\frac{\left.x^{2}-(a+2) x+a\right)}{x-2} & x \neq 2 \\ 2 & , x=2\end{array} \quad\right.$ is continuous at $x=2$, then the value of $a$ is
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Given, $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\mathrm{x}^{2}-(\mathrm{a}+2) \mathrm{x}+\mathrm{a}}{\mathrm{x}-2}, & \mathrm{x} \neq 2 \\ 2 & , x=2\end{array}\right.$
$$
\begin{aligned}
&\lim _{x \rightarrow 2} \frac{x^{2}-(a+2) x+a}{x-2} \\
&=\lim _{x \rightarrow 2} \frac{2 x-(a+2)}{1} \text { (by L. Hospital's rule) } \\
&=2 \times 2-a-2 \\
&=2-a
\end{aligned}
$$
Since, $f(x)$ is continuous at $x=2$, therefore
$\lim _{x \rightarrow 2} f(x)=f(2)$
$\begin{aligned} \therefore & 2-a &=2 \\ \Rightarrow & a &=0 . \end{aligned}$
$$
\begin{aligned}
&\lim _{x \rightarrow 2} \frac{x^{2}-(a+2) x+a}{x-2} \\
&=\lim _{x \rightarrow 2} \frac{2 x-(a+2)}{1} \text { (by L. Hospital's rule) } \\
&=2 \times 2-a-2 \\
&=2-a
\end{aligned}
$$
Since, $f(x)$ is continuous at $x=2$, therefore
$\lim _{x \rightarrow 2} f(x)=f(2)$
$\begin{aligned} \therefore & 2-a &=2 \\ \Rightarrow & a &=0 . \end{aligned}$
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