Question

let f(x) = \frac{x - 9}{|x - 9|}. find a) \lim_{x\to9^{+}}f(x), b) \lim_{x\to9^{-}}f(x), c) \lim_{x\to9}f(x), and d) f(9).
a) select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. \lim_{x\to9^{+}}f(x)= (simplify your answer.)
b. the limit does not exist.
b) select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. \lim_{x\to9^{-}}f(x)= (simplify your answer.)
b. the limit does not exist.
c) select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. \lim_{x\to9}f(x)= (simplify your answer.)
b. the limit does not exist.
d) select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. f(9)= (simplify your answer.)
b. f(9) does not exist.

Explanation:

Step1: Analyze right - hand limit ($x\to9^{+}$)

When $x\to9^{+}$, $x - 9>0$, so $|x - 9|=x - 9$. Then $f(x)=\frac{x - 9}{|x - 9|}=\frac{x - 9}{x - 9}=1$. So $\lim_{x\to9^{+}}f(x)=1$.

Step2: Analyze left - hand limit ($x\to9^{-}$)

When $x\to9^{-}$, $x - 9<0$, so $|x - 9|=-(x - 9)$. Then $f(x)=\frac{x - 9}{|x - 9|}=\frac{x - 9}{-(x - 9)}=-1$. So $\lim_{x\to9^{-}}f(x)=-1$.

Step3: Analyze two - sided limit ($x\to9$)

Since $\lim_{x\to9^{+}}f(x)=1$ and $\lim_{x\to9^{-}}f(x)=-1$, and $1
eq - 1$, the two - sided limit $\lim_{x\to9}f(x)$ does not exist.

Step4: Analyze $f(9)$

The function $f(x)=\frac{x - 9}{|x - 9|}$ is undefined at $x = 9$ because when $x = 9$, the denominator $|x - 9|=0$. So $f(9)$ does not exist.

Answer:

a) A. $\lim_{x\to9^{+}}f(x)=1$
b) A. $\lim_{x\to9^{-}}f(x)=-1$
c) B. The limit does not exist
d) B. $f(9)$ does not exist