Question

evaluate each limit algebraically. (a) $lim_{x
ightarrow3}\frac{x^{2}-9}{x^{2}+2x - 3}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-9=(x - 3)(x + 3)$ using the difference - of - squares formula $a^{2}-b^{2}=(a - b)(a + b)$. The denominator $x^{2}+2x - 3=(x + 3)(x - 1)$ by factoring the quadratic expression $ax^{2}+bx + c$ where $a = 1$, $b = 2$, $c=-3$ and finding two numbers that multiply to $ac=-3$ and add up to $b = 2$ (the numbers are 3 and - 1). So, $\lim_{x
ightarrow3}\frac{x^{2}-9}{x^{2}+2x - 3}=\lim_{x
ightarrow3}\frac{(x - 3)(x + 3)}{(x + 3)(x - 1)}$.

Step2: Cancel out the common factor

Since $x
eq3$ (we are taking the limit as $x$ approaches 3, not setting $x = 3$), we can cancel out the common factor $(x + 3)$ in the numerator and denominator. We get $\lim_{x
ightarrow3}\frac{x - 3}{x - 1}$.

Step3: Substitute $x = 3$

Substitute $x = 3$ into $\frac{x - 3}{x - 1}$. We have $\frac{3-3}{3 - 1}=\frac{0}{2}=0$.

Answer:

$0$