Hi,

$\displaystyle \lim_{\substack{x-3}} f(x)=\frac{x^2+4x-21}{x-3}$.

Using L'Hopital's rule (or algebra manipulation) we can determine the limit is 10. My question is, how is this possible if it is a rational function with a vertical asymptote at 3? Shouldn't it approach infinity?

After graphing it, I noticed that it happens to be a straight line and virtually equal to the graph of its derivative $\displaystyle f'(x)=2x+4$? This seems odd, as the graph said there was a hole at the point x=3, not a vertical asymptote. If I slightly change the function (such as x-4 in the denominator), it looks more like a rational function with a vertical asymptote.

I don't need any computational help with this problem - I just want to understand what's going on here.