Question

g(x) = \

$$\begin{cases}x & \\text{if }x < 1 \\\\ 4 & \\text{if }x = 1 \\\\ 2 - x^{2}& \\text{if }1 < x\\leq2 \\\\ x - 2& \\text{if }x > 2\\end{cases}$$

(a) evaluate each of the following, if it exists. (if an answer does not exist, enter dne.)
(i) \\(\lim_{x\to1^{-}}g(x)\\)
(ii) \\(\lim_{x\to1^{+}}g(x)\\)
(iii) g(1)
(iv) \\(\lim_{x\to2^{-}}g(x)\\)
(v) \\(\lim_{x\to2^{+}}g(x)\\)
(vi) \\(\lim_{x\to2}g(x)\\)
(b) sketch the graph of g.

Explanation:

Step1: Find left - hand limit as $x\to1$

For $x < 1$, $g(x)=x$. So, $\lim_{x\to1^{-}}g(x)=\lim_{x\to1^{-}}x = 1$.

Step2: Find right - hand limit as $x\to1$

For $1 < x\leq2$, $g(x)=2 - x^{2}$. So, $\lim_{x\to1^{+}}g(x)=\lim_{x\to1^{+}}(2 - x^{2})=2-1^{2}=1$.

Step3: Find $g(1)$

Given $g(1) = 4$.

Step4: Find left - hand limit as $x\to2$

For $1 < x\leq2$, $g(x)=2 - x^{2}$. So, $\lim_{x\to2^{-}}g(x)=\lim_{x\to2^{-}}(2 - x^{2})=2 - 2^{2}=-2$.

Step5: Find right - hand limit as $x\to2$

For $x>2$, $g(x)=x - 2$. So, $\lim_{x\to2^{+}}g(x)=\lim_{x\to2^{+}}(x - 2)=2 - 2=0$.

Step6: Find $\lim_{x\to2}g(x)$

Since $\lim_{x\to2^{-}}g(x)
eq\lim_{x\to2^{+}}g(x)$, $\lim_{x\to2}g(x)$ does not exist (DNE).

Answer:

(i) 1
(ii) 1
(iii) 4
(iv) - 2
(v) 0
(vi) DNE