A function $f(x)$ is said to have a removable discontinuity at $x=a$ if:
1. $f$ is either not defined or not continuous at $x=a$.
2. $f(a)$ could either be defined or redefined so that the new function is continuous at $x=a$.

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Let $f(x) = \begin{cases} \frac{7}{x}+\frac{-6 x+14}{x(x-2)}, &\text{if}\ x\neq0,2\\ 2, &\text{if}\ x=0 \end{cases}$
Show that $f(x)$ has a removable discontinuity at $x=0$ and determine what value for $f(0)$ would make $f(x)$ continuous at $x=0$.
Must redefine $f(0)=$.
Hint: Try combining the fractions and simplifying.
The discontinuity at $x=2$ is not a removable discontinuity, just in case you were wondering.