10. $lim_{x \to 2}\frac{x^{2}-4x + 4}{x^{2}+x - 6}$
10. $lim_{x \to 2}\frac{x^{2}-4x + 4}{x^{2}+x - 6}$
Answer
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 $\lim_{x\rightarrow2}\frac{(x - 2)^{2}}{(x - 2)(x + 3)}$.
Step2: Simplify the function
Cancel out the common factor $(x - 2)$ (since $x\neq2$ when taking the limit), we get $\lim_{x\rightarrow2}\frac{x - 2}{x + 3}$.
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:
$0$