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

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.

• If the numerator is approaching a number that is clearly non-zero, while the denominator is going to zero, that would indicate that your function must be heading toward somewhere bad. What is the sign going to be as you approach your limit point from either side? Does it change? Now in this case, it might help to notice your function is $\frac {(x-2)^2}{(x-3)^2}.$ By the way the "$\frac 1x$ trick" really only works if $x$ is going to infinity. – Doug M Jan 23 '18 at 19:05

$$\lim_{x \to 3} \frac{x^2-4x+4}{x^2-6x+9}=\lim_{x \to 3} \frac{(x-2)^2}{(x-3)^2} > 0$$
use that $$x^2-4x+4=(x-2)^2$$ and $$x^2-6x+9=(x-3)^2$$
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 $\frac{1}{x}$).
\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 $\lim_{x\rightarrow 3}\dfrac{(x-2)^{2}}{(x-3)^{2}}=\infty$.