Question

let $g(x)=\frac{x^{2}+x - 2}{|x - 1|}$.
(a) find the following limits. (if an answer does not exist, enter dne.)
(i) $lim_{x
ightarrow1^{+}}g(x)$
(ii) $lim_{x
ightarrow1^{-}}g(x)$
(b) does $lim_{x
ightarrow1}g(x)$ exist?
yes
no
(c) sketch the graph of $g$.

Explanation:

Step1: Factor the numerator

First, factor \(x^{2}+x - 2=(x + 2)(x - 1)\). So \(g(x)=\frac{(x + 2)(x - 1)}{|x - 1|}\).

Step2: Analyze the right - hand limit (\(x\to1^{+}\))

When \(x\to1^{+}\), \(x-1>0\), so \(|x - 1|=x - 1\). Then \(\lim_{x\to1^{+}}g(x)=\lim_{x\to1^{+}}\frac{(x + 2)(x - 1)}{x - 1}=\lim_{x\to1^{+}}(x + 2)\). Substitute \(x = 1\) into \(x + 2\), we get \(\lim_{x\to1^{+}}(x + 2)=3\).

Step3: Analyze the left - hand limit (\(x\to1^{-}\))

When \(x\to1^{-}\), \(x - 1<0\), so \(|x - 1|=-(x - 1)\). Then \(\lim_{x\to1^{-}}g(x)=\lim_{x\to1^{-}}\frac{(x + 2)(x - 1)}{-(x - 1)}=\lim_{x\to1^{-}}-(x + 2)\). Substitute \(x = 1\) into \(-(x + 2)\), we get \(\lim_{x\to1^{-}}-(x + 2)=-3\).

Step4: Determine if the limit exists

Since \(\lim_{x\to1^{+}}g(x)=3\) and \(\lim_{x\to1^{-}}g(x)=-3\), and \(3
eq - 3\), \(\lim_{x\to1}g(x)\) does not exist.

Answer:

(i) 3
(ii) -3
(b) No
(c) To sketch the graph:

• For \(x>1\), \(g(x)=x + 2\), which is a straight - line with slope \(m = 1\) and \(y\) - intercept \(y=2\) (but we consider the part \(x>1\)).
• For \(x<1\), \(g(x)=-(x + 2)=-x - 2\), which is a straight - line with slope \(m=-1\) and \(y\) - intercept \(y = - 2\) (but we consider the part \(x<1\)). There is a break at \(x = 1\) since the left - hand and right - hand limits are different.