#### Answer

To show that $\lim_{x\to2}h(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-2|\lt\delta\Rightarrow|h(x)-M|\ge\epsilon$$

#### Work Step by Step

To show that $\lim_{x\to2}h(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-2|\lt\delta\Rightarrow|h(x)-M|\ge\epsilon$$
a) $\lim_{x\to2}h(x)\ne 4$
Let $\epsilon=1$
- For $x\gt2$, $h(x)=2$, meaning that $$|h(x)-4|=|2-4|=|-2|=2\gt1$$
That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne4$
b) Prove that $\lim_{x\to2}h(x)\ne3$
Let $\epsilon=1/2$
Again, for $x\gt2$, $h(x)=2$, meaning that $$|h(x)-3|=|2-3|=|-1|=1\gt\frac{1}{2}$$
That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne3$
c) Prove that $\lim_{x\to2}h(x)\ne2$
Let $\epsilon=0.2$
- For $x\lt2$, $h(x)=x^2$, meaning that $|h(x)-2|=|x^2-2|$
- Since $0\lt|x-2|\lt\delta$, that means $-\delta\lt x-2\lt\delta$, and $2-\delta\lt x\lt2+\delta$
Now we need to take a value of $x$ so that $x\lt2$ and $2-\delta\lt x\lt2+\delta$. I would take $x=2-(\delta/2)$, thus $x^2=(2-(\delta/2))^2=4-2\delta+(\delta^2/4)$
That means $$|h(x)-2|=|x^2-2|=|4-2\delta+(\delta^2/4)-2|=|2-2\delta+(\delta^2/4)|$$
The graph of the function $|h(x)-2|=|2-2\delta+(\delta^2/4)|$ is shown below.
As we can see from the graph, as $\delta\ge1$, $|h(x)-2|\ge0.25\gt0.2$
That completes our proof. We can conclude now that $\lim_{x\to2}h(x)\ne2$