$$ lim _ { x \rightarrow 3 } \frac { x ^ { 2 } - 6 x + 9 } { x ^ { 2 } - 9 } = $$
$$ lim _ { x \rightarrow 3 } \frac { x ^ { 2 } - 6 x + 9 } { x ^ { 2 } - 9 } = $$
Answer
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\rightarrow3}\frac{x^{2}-6x + 9}{x^{2}-9}=\lim_{x\rightarrow3}\frac{(x - 3)^{2}}{(x + 3)(x - 3)})
Step2: Simplify the expression
Cancel out the common factor ((x - 3)) (since (x\neq3) when taking the limit), we get (\lim_{x\rightarrow3}\frac{x - 3}{x + 3})
Step3: Substitute (x = 3) into the simplified expression
(\frac{3-3}{3 + 3}=\frac{0}{6}=0)
Answer:
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