One-sided infinite limit as x -> 2+

Factor the denominator: $x^2-5x+6=(x-2)(x-3)$. As $x \to 2^+$, take $x$ slightly larger than 2, such as $x=2.1$. Then $x-2 = 0.1 > 0$ (a small positive number) and $x-3 = -0.9 < 0$ (a negative number close to $-1$).

Multiplying these factors gives a small negative number, so the denominator approaches $0$ from the negative side. Since the numerator is $1$ (positive), the fraction $\frac{1}{(x-2)(x-3)}$ becomes a large negative number, so $\lim_{x\to 2^+} f(x) = -\infty$.

Using the factorization $x^2-5x+6=(x-2)(x-3)$, take $x$ slightly less than 3, such as $x=2.9$. Then $x-2 = 0.9 > 0$ and $x-3 = -0.1 < 0$ (a small negative number).

The product $(x-2)(x-3) = (0.9)(-0.1) = -0.09$ is a small negative number, so the denominator approaches $0$ from the negative side. Since the numerator is $1$, we have $\frac{1}{(x-2)(x-3)} \to -\infty$, giving $\lim_{x\to 3^-} f(x) = -\infty$.