Answer
(a) Let $\epsilon=0.5$ then prove that for $x\lt-1$, $|g(x)-2|\ge\epsilon$.
(b) $\lim_{x\to-1}g(x)$ does appear to exist, and it equals $1$.
Work Step by Step
(a) $\lim_{x\to-1}g(x)\ne2$
We need to choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-(-1)|\lt\delta\Rightarrow|g(x)-2|\ge\epsilon$$
- Let $\epsilon=0.5$
- Looking at the graph, for $x\lt-1$, $g(x)\lt-1$
This means $g(x)-2\lt-1-2=-3$
Therefore, $|g(x)-2|\gt3\gt0.5$
Hence, $\lim_{x\to-1}g(x)\ne2$ according to limit definition.
(b) $\lim_{x\to-1}g(x)$ does appear to exist, and it equals $1$.
Because if we choose an arbitrary $\epsilon\gt0$ and let the value of $g(x)$ fluctuate between $1-\epsilon$ and $1+\epsilon$, we can always find a corresponding value of $\delta\gt0$ and the corresponding values of $x$ would be restricted in the interval $(-1-\delta,-1+\delta)$.
In other words, $$0\lt|x-(-1)|\lt\delta\Rightarrow|g(x)-1|\lt\epsilon$$
