Question

$$ lim _ { x ightarrow 3 } \frac { x ^ { 2 } - 6 x + 9 } { x ^ { 2 } - 9 } = $$

Explanation:

Step1: Factor numerator and denominator

Numerator: \(x^{2}-6x + 9=(x - 3)^{2}\)
Denominator: \(x^{2}-9=(x + 3)(x - 3)\)
So, \(\lim_{x
ightarrow3}\frac{x^{2}-6x + 9}{x^{2}-9}=\lim_{x
ightarrow3}\frac{(x - 3)^{2}}{(x + 3)(x - 3)}\)

Step2: Simplify the expression

Cancel out the common factor \((x - 3)\) (since \(x
eq3\) when taking the limit), we get \(\lim_{x
ightarrow3}\frac{x - 3}{x + 3}\)

Step3: Substitute \(x = 3\) into the simplified expression

\(\frac{3-3}{3 + 3}=\frac{0}{6}=0\)

Answer:

\(0\)