Question

enter each answer as a whole number (like -2, -5, 626, 73772), or a fraction (like 35/4983) or dne for undefined (does not exist).
lim_{x→2^+} (f(x) - 4)/f(x + 1)=
lim_{x→2^-} f(f(x)+4)=
lim_{h→0} (f(2 + h)-f(2))/h=

Explanation:

Step1: Analyze $\lim_{x

ightarrow2^{+}}\frac{f(x)-4}{f(x + 1)}$
As $x
ightarrow2^{+}$, from the graph, $f(x)
ightarrow3$ and $f(x + 1)$ when $x
ightarrow2^{+}$ means considering $f(3^{+})$ and $f(3^{+})=3$. So $\lim_{x
ightarrow2^{+}}\frac{f(x)-4}{f(x + 1)}=\frac{3 - 4}{3}=-\frac{1}{3}$.

Step2: Analyze $\lim_{x

ightarrow2^{-}}f(f(x)+4)$
As $x
ightarrow2^{-}$, $f(x)
ightarrow3$. Then $f(x)+4
ightarrow7$. From the graph, $f(7)=6$. So $\lim_{x
ightarrow2^{-}}f(f(x)+4)=6$.

Step3: Analyze $\lim_{h

ightarrow0}\frac{f(2 + h)-f(2)}{h}$
This is the definition of the derivative at $x = 2$. The left - hand limit of the slope as $h
ightarrow0^{-}$ and the right - hand limit of the slope as $h
ightarrow0^{+}$ are not equal (the graph has a corner at $x = 2$). So $\lim_{h
ightarrow0}\frac{f(2 + h)-f(2)}{h}$ DNE.

Answer:

$-\frac{1}{3}$
$6$
DNE