Question

enter each answer as a whole number (like -2, -5, 626, 73772), or a fraction (like $\frac{35}{4983}$) or dne for undefined (does not exist).
$lim_{x
ightarrow2^{-}}\frac{f(x)-4}{f(x + 1)}=
lim_{x
ightarrow2^{-}}f(f(x)+4)=
lim_{h
ightarrow0}\frac{f(2 + h)-f(2)}{h}=$

Explanation:

Step1: Analyze limits from the graph

First, find $f(x)$ values near $x = 2$. As $x\to2^{-}$, $f(x)\to3$. Also, when $x\to2^{-}$, $f(x + 1)$: let $t=x + 1$, when $x\to2^{-}$, $t\to3^{-}$ and $f(t)\to3$.

Step2: Calculate first - limit

For $\lim_{x\to2^{-}}\frac{f(x)-4}{f(x + 1)}$, substitute $f(x)\to3$ and $f(x + 1)\to3$ as $x\to2^{-}$. We get $\frac{3 - 4}{3}=\frac{-1}{3}$.

Step3: Analyze inner - function for second limit

As $x\to2^{-}$, $f(x)\to3$. Then $f(x)+4\to3 + 4=7$. And $f(f(x)+4)$: when the input is $7$, from the graph $f(7)=6$. So $\lim_{x\to2^{-}}f(f(x)+4)=6$.

Step4: Analyze derivative - like limit

The limit $\lim_{h\to0}\frac{f(2 + h)-f(2)}{h}$ represents the left - hand derivative at $x = 2$. The function is linear on the left - hand side of $x = 2$. The slope of the line segment from $(1,1)$ to $(3,3)$ is $m=\frac{3 - 1}{3 - 1}=1$.

Answer:

$\lim_{x\to2^{-}}\frac{f(x)-4}{f(x + 1)}=-\frac{1}{3}$
$\lim_{x\to2^{-}}f(f(x)+4)=6$
$\lim_{h\to0}\frac{f(2 + h)-f(2)}{h}=1$