<s>[INST] For what value c is $f(x) = \frac{x^2+5x-14}{x^3-x^2-x-2}$ at 2 continuous? Im pretty new to continuinity and it seems for me the function can as easily be repaired with $c = 2$ so that $g(x) = \begin{cases}
\frac{x^2+5x-14}{x^3-x^2-x-2}, & x > 2\\
0, & x = 2 \\ \frac{x^2+5x-14}{x^3-x^2-x-2}, & x < 2
\end{cases} $
, because $f(2) = \frac{0}{0}$, which is a gap in the function and the function. But plotting the function proves it otherwise, because the value is ~1,5. So I'd have to take the limes from the left and right to this point, $left-\lim\limits_{n \to 2} \frac{x^2+5x-14}{x^3-x^2-x-2} = \frac{2^2+5\cdot2-14}{2^3-2^2-2-2} = \frac{0}{0}$... Help is appreciated!
 [/INST] Instead of plugging $2$ directly into the expression, first factor the numerator and denominator as
$$f(x) = \frac{x^2 + 5x - 14}{x^3 - x^2 - x - 2} = \frac{(x + 7)(x - 2)}{(x - 2)(x^2 + x + 1)} = \frac{x + 7}{x^2 + x + 1}.$$
Then substituting $x = 2$ tells you that
$$\lim_{x \to 2}f(x) = \frac{2 + 7}{2^2 + 2 + 1} = \frac{9}{7}.$$
So the function is continuous if $g(c) = 9/7$.
</s>