Question

b. evaluate $limlimits_{x \to -3} \frac{x + 3}{x^{2} - 9}$

Explanation:

Step1: Factor the denominator

We know that \(x^{2}-9=(x + 3)(x - 3)\) (using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b = 3\)). So the limit becomes \(\lim_{x
ightarrow - 3}\frac{x + 3}{(x + 3)(x - 3)}\).

Step2: Simplify the function

Cancel out the common factor \((x + 3)\) (for \(x
eq - 3\), when taking the limit as \(x
ightarrow - 3\), \(x
eq - 3\) in the neighborhood around \(x=-3\) except at \(x =-3\)). The function simplifies to \(\lim_{x
ightarrow - 3}\frac{1}{x - 3}\).

Step3: Substitute \(x=-3\)

Substitute \(x=-3\) into \(\frac{1}{x - 3}\). We get \(\frac{1}{-3-3}\).

Answer:

\(-\frac{1}{6}\)