Question

1. $lim_{x

ightarrow1^{+}}\frac{x^{2}-5x + 6}{x - 1}$

Explanation:

Step1: Factor the numerator

First, factor $x^{2}-5x + 6=(x - 2)(x - 3)$. So the limit becomes $\lim_{x
ightarrow1^{+}}\frac{(x - 2)(x - 3)}{x - 1}$.

Step2: Analyze the behavior as $x

ightarrow1^{+}$
As $x
ightarrow1^{+}$, $(x - 2)
ightarrow - 1$, $(x - 3)
ightarrow - 2$, and $(x - 1)
ightarrow0^{+}$.
The product of $(x - 2)$ and $(x - 3)$ is $(-1)\times(-2)=2$. And since the denominator $x - 1$ approaches $0$ from the positive - side, $\frac{2}{x - 1}
ightarrow+\infty$.

Answer:

$+\infty$