The function $f$ is defined by the formula
$\qquad f(x) =
\begin{cases}
\dfrac{\dfrac{1}{x-2}-1}{x-3} &\text{if $x < 3$}\\\\
cx+1 &\text{if $x\geq3 $}
\end{cases}$
**What is the value of $c$ that will make $f$ continuous at $x=3$ ?**
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