Determine to which side of infinity the limit goes without looking at the graph

Question

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.

Answer 1

$$\lim_{x \to 3} \frac{x^2-4x+4}{x^2-6x+9}=\lim_{x \to 3} \frac{(x-2)^2}{(x-3)^2} > 0$$

Answer 2

use that $$x^2-4x+4=(x-2)^2$$ and $$x^2-6x+9=(x-3)^2$$

Answer 3

Hint : use a factorised expression and compute the sign ($+$ or $-$) of each factor. This has to be done on every part that vanishes at $x=3$ (eg the denominator as you already know). For those parts that don't vanish (like the numerator) you don't have to do that, simply compute the sign of its value at $x=3$.

The trick that you are using is mostly useful when you look at the limits as $x$ tends to infinity (but then you would have to multiply by $\frac{1}{x^2}$ instead of $\frac{1}{x}$).

Answer 4

\begin{align*} \dfrac{x^{2}-4x+4}{x^{2}-6x+9}&=\dfrac{(x-2)^{2}}{(x-3)^{2}}, \end{align*} where for sufficiently closed $x$ to $3$, we have $|x-3|<\dfrac{1}{2}$, then $|x-2|\geq 1-|x-3|>\dfrac{1}{2}$, so $(x-2)^{2}\geq\dfrac{1}{4}$ for all such $x$, then \begin{align*} \dfrac{(x-2)^{2}}{(x-3)^{2}}\geq\dfrac{1}{4}\dfrac{1}{(x-3)^{2}}, \end{align*} where $\lim_{x\rightarrow 3}\dfrac{1}{(x-3)^{2}}=\infty$ is easy to see, so $\lim_{x\rightarrow 3}\dfrac{(x-2)^{2}}{(x-3)^{2}}=\infty$.