Question

given the following function: $f(x)=\begin{cases}4x - c& \text{if }xleq4,\\3+(x - 4)^2sinleft(\frac{1}{x - 4}
ight)&\text{if }x>4.end{cases}$ (a) which of the following is true? if $c = 100$, the one - sided limits at $x = 4$ both exist, but do not match. there is no value of $c$ that makes $f$ continuous because one of the one - sided limits at $x = 4$ does not exist. there is no value of $c$ that makes $f$ continuous at $x = 4$ because $(x - 4)^2sinleft(\frac{1}{x - 4}
ight)$ is not defined at $x = 4$. select all possible options that apply. (b) find the value of $c$ that makes the previous function continuous. if no such value of $c$ exists, enter dne. $c=$ symbolic expression

Explanation:

Step1: Calculate left - hand limit

For $x\leq4$, $f(x)=4x - c$. The left - hand limit as $x\to4^{-}$ is $\lim_{x\to4^{-}}(4x - c)=4\times4 - c=16 - c$.

Step2: Calculate right - hand limit

For $x > 4$, $f(x)=3+(x - 4)^{2}\sin(\frac{1}{x - 4})$. We know that $- 1\leq\sin(\frac{1}{x - 4})\leq1$. Then $-(x - 4)^{2}\leq(x - 4)^{2}\sin(\frac{1}{x - 4})\leq(x - 4)^{2}$. By the Squeeze Theorem, $\lim_{x\to4^{+}}(x - 4)^{2}\sin(\frac{1}{x - 4}) = 0$. So $\lim_{x\to4^{+}}f(x)=\lim_{x\to4^{+}}(3+(x - 4)^{2}\sin(\frac{1}{x - 4}))=3+0 = 3$.

Step3: Analyze part (a)

If $c = 100$, $\lim_{x\to4^{-}}f(x)=16 - 100=-84$ and $\lim_{x\to4^{+}}f(x)=3$, so the one - sided limits at $x = 4$ both exist but do not match. The function $f(x)$ is continuous at $x = 4$ if $\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{+}}f(x)$. The right - hand limit exists. The fact that $(x - 4)^{2}\sin(\frac{1}{x - 4})$ is not defined at $x = 4$ is not the reason for non - continuity since we are taking the limit as $x\to4$. The correct statement for part (a) is: If $c = 100$, the one sided limits at $x = 4$ both exist, but do not match.

Step4: Solve for $c$ in part (b)

For $f(x)$ to be continuous at $x = 4$, we set $\lim_{x\to4^{-}}f(x)=\lim_{x\to4^{+}}f(x)$. So $16 - c=3$. Solving for $c$, we get $c = 13$.

Answer:

(a) If $c = 100$, the one sided limits at $x = 4$ both exist, but do not match.
(b) $c = 13$