Question

find the limits in a) through c) below for the function f(x) = \frac{x^{2}-6x - 36}{x - 9}. use -\infty and \infty when appropriate.
a) select the correct choice below and fill in any answer boxes in your choice.
a. \lim_{x\to9^{-}}f(x)=13 (simplify your answer.)
b. the limit does not exist and is neither -\infty nor \infty.
b) select the correct choice below and fill in any answer boxes in your choice.
a. \lim_{x\to9^{+}}f(x)= (simplify your answer.)
b. the limit does not exist and is neither -\infty nor \infty.

Explanation:

Step1: Factor the numerator

First, factor the numerator $x^{2}-5x - 36$. We know that $x^{2}-5x - 36=(x - 9)(x+4)$ since $(-9)\times4=-36$ and $-9 + 4=-5$. So the function $f(x)=\frac{x^{2}-5x - 36}{x - 9}=\frac{(x - 9)(x + 4)}{x - 9}$. For $x
eq9$, we can cancel out the $(x - 9)$ terms, and $f(x)=x + 4$.

Step2: Find the right - hand limit

To find $\lim_{x
ightarrow9^{+}}f(x)$, since $f(x)=x + 4$ for $x
eq9$, we substitute $x = 9$ into $x+4$. So $\lim_{x
ightarrow9^{+}}f(x)=\lim_{x
ightarrow9^{+}}(x + 4)=9+4=13$.

Answer:

a) A. $\lim_{x
ightarrow9^{-}}f(x)=13$
b) A. $\lim_{x
ightarrow9^{+}}f(x)=13$