Question

question 16
let $f(x)=\

$$\begin{cases}-1x - 4&\\text{if }x < 6\\\\cx&\\text{if }x > 6\\end{cases}$$

\lim\limits_{x\to 6^{-}}f(x)=\square
\lim\limits_{x\to 6^{+}}f(x)=\square
if \lim\limits_{x\to 6}f(x) exists, then it must be \square (enter
this means $c = \square
question help: \boxed{video} \boxed{written example}

Explanation:

Step1: Compute left-hand limit

Use $f(x)=-x-4$ for $x<6$.
$\lim_{x \to 6^-} f(x) = -(6) - 4$

Step2: Calculate left-hand limit value

Simplify the expression.
$\lim_{x \to 6^-} f(x) = -6 - 4 = -10$

Step3: Compute right-hand limit

Use $f(x)=cx$ for $x>6$.
$\lim_{x \to 6^+} f(x) = c(6) = 6c$

Step4: Set limits equal for existence

For $\lim_{x \to 6} f(x)$ to exist, left and right limits must be equal.
$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x) \implies -10 = 6c$

Step5: Solve for $c$

Rearrange to isolate $c$.
$c = \frac{-10}{6} = -\frac{5}{3}$

Answer:

$\lim_{x \to 6^-} f(x) = -10$
$\lim_{x \to 6^+} f(x) = 6c$
If $\lim_{x \to 6} f(x)$ exists, then it must be $-10$
$c = -\frac{5}{3}$