Question

let (f(x)=\begin{cases}-\frac{21}{x - 2}&\text{if }x<5\\-32&\text{if }x = 5\\frac{7}{x - 6}&\text{if }x>5end{cases}
compute the quantities below. write \dne\ if the limit does not exist or the value is undefined.
(lim_{x
ightarrow5^{-}}f(x)=)
(lim_{x
ightarrow5^{+}}f(x)=)
(f(5)=)
since the above three quantities are select an answer, we know that (f) is select an answer at (x = 5).
list all numbers at which (f) is discontinuous. be sure to check the functions defined to the left and right of 5 for discontinuities.
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Explanation:

Step1: Find left - hand limit

We use the function definition for $x < 5$, which is $f(x)=-\frac{21}{x - 2}$. To find $\lim_{x
ightarrow5^{-}}f(x)$, we substitute $x = 5$ into $-\frac{21}{x - 2}$. So, $\lim_{x
ightarrow5^{-}}f(x)=-\frac{21}{5 - 2}=-\frac{21}{3}=- 7$.

Step2: Find right - hand limit

We use the function definition for $x>5$, which is $f(x)=\frac{7}{x - 6}$. To find $\lim_{x
ightarrow5^{+}}f(x)$, we substitute $x = 5$ into $\frac{7}{x - 6}$. So, $\lim_{x
ightarrow5^{+}}f(x)=\frac{7}{5 - 6}=-7$.

Step3: Find the function value at $x = 5$

From the function definition, when $x = 5$, $f(5)=-32$.

Step4: Determine continuity

Since $\lim_{x
ightarrow5^{-}}f(x)=-7$, $\lim_{x
ightarrow5^{+}}f(x)=-7$ and $f(5)=-32$, the three quantities are not equal. A function $y = f(x)$ is continuous at $x=a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. So $f(x)$ is discontinuous at $x = 5$. Also, the function $y=-\frac{21}{x - 2}$ is undefined at $x = 2$ and $y=\frac{7}{x - 6}$ is undefined at $x = 6$.

Answer:

$\lim_{x
ightarrow5^{-}}f(x)=-7$
$\lim_{x
ightarrow5^{+}}f(x)=-7$
$f(5)=-32$
Since the above three quantities are not equal, we know that $f$ is discontinuous at $x = 5$.
List ALL numbers at which $f$ is discontinuous: $2,5,6$