Question

$lim_{x \to 9}\frac{x^{2}-4x - 45}{x^{2}-81}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-4x - 45=(x - 9)(x+5)$ and the denominator $x^{2}-81=(x - 9)(x + 9)$. So the function becomes $\frac{(x - 9)(x + 5)}{(x - 9)(x + 9)}$.

Step2: Simplify the function

Cancel out the common factor $(x - 9)$ (since $x
eq9$ when taking the limit), we get $\frac{x + 5}{x + 9}$.

Step3: Substitute $x = 9$ into the simplified function

Substitute $x=9$ into $\frac{x + 5}{x + 9}$, we have $\frac{9+5}{9 + 9}=\frac{14}{18}=\frac{7}{9}$.

Answer:

$\frac{7}{9}$