Question

let $h(x)=\begin{cases}x & \text{if }x < 0\\x^{2}&\text{if }0 < xleq2\\8 - x&\text{if }x>2end{cases}$
find each of the following limits. if the limit does not exist, enter dne below.
(a) $lim_{x
ightarrow0^{-}}h(x)=$
(b) $lim_{x
ightarrow0}h(x)=$
(c) $lim_{x
ightarrow1}h(x)=$
(d) $lim_{x
ightarrow2^{-}}h(x)=$
(e) $lim_{x
ightarrow2^{+}}h(x)=$
(f) $lim_{x
ightarrow2}h(x)=$

Explanation:

Step1: Analyze left - hand limit as $x\to0^{-}$

When $x\to0^{-}$, we use the part of the function $h(x)=x$ for $x < 0$. So, $\lim_{x\to0^{-}}h(x)=\lim_{x\to0^{-}}x = 0$.

Step2: Analyze right - hand limit as $x\to0^{+}$

When $x\to0^{+}$, we use the part of the function $h(x)=x^{2}$ for $0 < x\leq2$. So, $\lim_{x\to0^{+}}h(x)=\lim_{x\to0^{+}}x^{2}=0$.

Step3: Analyze limit as $x\to1$

When $x\to1$, since $0 < 1\leq2$, we use $h(x)=x^{2}$. So, $\lim_{x\to1}h(x)=\lim_{x\to1}x^{2}=1$.

Step4: Analyze left - hand limit as $x\to2^{-}$

When $x\to2^{-}$, since $0 < x\leq2$, we use $h(x)=x^{2}$. So, $\lim_{x\to2^{-}}h(x)=\lim_{x\to2^{-}}x^{2}=4$.

Step5: Analyze right - hand limit as $x\to2^{+}$

When $x\to2^{+}$, since $x > 2$, we use $h(x)=8 - x$. So, $\lim_{x\to2^{+}}h(x)=\lim_{x\to2^{+}}(8 - x)=6$.

Step6: Analyze limit as $x\to2$

Since $\lim_{x\to2^{-}}h(x)=4$ and $\lim_{x\to2^{+}}h(x)=6$, $\lim_{x\to2}h(x)$ does not exist (DNE).

Answer:

(a) $0$
(b) $0$
(c) $1$
(d) $4$
(e) $6$
(f) DNE