Question

question 2
0/1 pt 3 details
enter each answer as a whole number (like -5, -9, 579, 53963), or a fraction (like \frac{44}{5270}) or dne for undefined (does not exist).
\lim_{x \to 0^{+}}\frac{f(x)-4}{f(x + 4)}=
\lim_{x \to 2^{-}}f(f(x)+4)=
\lim_{h \to 0}\frac{f(2 + h)-f(2)}{h}=
check answer

Explanation:

Step1: Analyze $\lim_{x\to0^{+}}\frac{f(x)-4}{f(x + 4)}$

As $x\to0^{+}$, from the graph, $f(x)\to4$ and $f(x + 4)\to3$. So, $\lim_{x\to0^{+}}\frac{f(x)-4}{f(x + 4)}=\frac{4 - 4}{3}=0$.

Step2: Analyze $\lim_{x\to2^{-}}f(f(x)+4)$

As $x\to2^{-}$, $f(x)\to3$. Then $f(x)+4\to7$. And $f(7) = 6$. So, $\lim_{x\to2^{-}}f(f(x)+4)=6$.

Step3: Analyze $\lim_{h\to0}\frac{f(2 + h)-f(2)}{h}$

This is the definition of the derivative at $x = 2$. The left - hand slope at $x=2$ is $\frac{3 - 1}{2-1}=2$, the right - hand slope at $x = 2$ is $\frac{3-2}{3 - 2}=1$. Since the left - hand and right - hand slopes are not equal, the derivative does not exist. So, $\lim_{h\to0}\frac{f(2 + h)-f(2)}{h}=\text{DNE}$.

Answer:

0
6
DNE