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Let $f(x)=\left\{\begin{array}{rr}\frac{x^3+x^2-16 x+20}{(x-2)^2}, & \text { if } x \neq 2 \\ k, & \text { if } x=2\end{array}\right.$. If $f(x)$ be continuous for all $x$, then $k=$
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The correct answer is:
$7$
For continuous $\lim _{x \rightarrow 2} f(x)=f(2)=k$
$\begin{aligned} & \Rightarrow k=\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4 x+4\right)(x+5)}{(x-2)^2}=7\end{aligned}$
$\begin{aligned} & \Rightarrow k=\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4 x+4\right)(x+5)}{(x-2)^2}=7\end{aligned}$
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