g(x) = \\begin{cases}x & \\text{if }x < 1 \\\\ 4 & \\text{if }x = 1 \\\\ 2 - x^{2}& \\text{if }1 < x\\leq2…

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

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

Answer

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

Answer:

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