What is wrong with the application of L'Hopital's rule to $\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$? The question is
$$\lim_{x\rightarrow 2}\frac{x^2 - 7x + 10}{x^2 + x - 6}$$  If we plug in $x = 2$ we see that both quadratrics have a zero there.  "Logically" this is a L'Hopital question, so if I take the derivative of top and bottom and find the limit I get:
$$\lim_{x\rightarrow2}\frac{2x-7}{2x+1} = \frac{-3}{5}$$
However, if I plot the function as close to $x = 2$ that I wish,, I see that the right handed limit and the left handed limit are positive and negative infinity, respectively, and of course my plotting software cannot plot the $x = 2$ case.
This concerns me.  On the one hand it seems that the limit should not exist, but on the other hand L'Hopital pumps out an answer.
What did I miss?

It holds $x^2-7x+10=(x-2)(x-5)$ and $x^2+x-6=(x-2)(x+3)$. Hence, we have $$\frac{x^2-7x+10}{x^2+x-6}=\frac{(x-5)(x-2)}{(x+3)(x-2)}=\frac{x-5}{x+3}$$
for all $x\notin\{-3,2\}$. Hence, the pole $x=2$ is removable, because we can extend the domain of the lefthandside by $x=2$, such that the function is still continuous. So your result $-\frac 35$ is correct, because $\frac{2-5}{2+3}=-\frac 35$.
