Question

1. $lim_{x \to 2}\frac{x^{2}-4x + 4}{x^{2}+x - 6}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-4x + 4=(x - 2)^{2}$, and the denominator $x^{2}+x - 6=(x - 2)(x+3)$. So the function becomes $\lim_{x
ightarrow2}\frac{(x - 2)^{2}}{(x - 2)(x + 3)}$.

Step2: Simplify the function

Cancel out the common factor $(x - 2)$ (since $x
eq2$ when taking the limit), we get $\lim_{x
ightarrow2}\frac{x - 2}{x + 3}$.

Step3: Substitute $x = 2$

Substitute $x=2$ into $\frac{x - 2}{x + 3}$, we have $\frac{2-2}{2 + 3}=\frac{0}{5}=0$.

Answer:

$0$