Today I got the following limit:
$$\lim_{x \to 3} \frac{x^2-4x+4}{x^2-6x+9}$$ I used the multiply by $\frac{\frac{1}{x}}{\frac{1}{x}}$ trick to reach the following: $$ \frac{x^2-4x+4}{x^2-6x+9} * \frac{\frac{1}{x}}{\frac{1}{x}} = \frac{x-4+\frac{4}{x}}{x-6+\frac{9}{x}}$$ Plugging in 3 gives $\frac{\frac{1}{3}}{0}$ which lets me know it's approaching $+\infty$ or $-\infty$.
I know from the answers and from plotting a graph that the final answer equals $+\infty$ but I don't see how I would be able to calculate this without either looking at the graph, or plugging in values very close to 3. We also didn't learn about l'Hôpital yet.
So the question: how do I know if the answer is $+\infty$, $-\infty$ or that the limit doesn't exist at all without using a calculator?
I feel like I'm missing a basic piece of knowledge here that will give me that "ooohh that's how it works" factor.