Answer
For all these exercises, to prove $\lim_{x\to1}f(x)\ne M$, let $\epsilon=1/2$ and show that for each $\delta\gt0$, there is a value of $x$ such that $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-M|\ge1/2$$
Work Step by Step
$f(x)=x$ for $x\lt1$ and $f(x)=x+1$ for $x\gt1$
a) $\epsilon=1/2$
As we were told, we need to show that for each $\delta\gt0$ there exists a value of $x$ such that $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-2|\ge\frac{1}{2}$$
- For $x\lt1$, $f(x)=x$, meaning that $|f(x)-2|=|x-2|$
- Since $0\lt|x-1|\lt\delta$, that means $-\delta\lt x-1\lt\delta$, and $1-\delta\lt x\lt1+\delta$
Now we need to take a value of $x$ so that $x\lt1$ and $1-\delta\lt x\lt1+\delta$. I would take $x=1-(\delta/4)$
That means $$|f(x)-2|=|x-2|=|1-\frac{\delta}{4}-2|=|-1-\frac{\delta}{4}|=1+\frac{\delta}{4}\gt\frac{1}{2}$$ (since $\delta\gt0$)
That completes our proof. We can conclude now that $\lim_{x\to1}f(x)\ne2$
b) Prove that $\lim_{x\to1}f(x)\ne1$
Let $\epsilon=1/2$
Again, we need to show that for each $\delta\gt0$ there exists a value of $x$ such that $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\ge\frac{1}{2}$$
- For $x\gt1$, $f(x)=x+1$, meaning that $|f(x)-1|=|x+1-1|=|x|$
- Since $0\lt|x-1|\lt\delta$, that means $-\delta\lt x-1\lt\delta$, and $1-\delta\lt x\lt1+\delta$
Now we need to take a value of $x$ so that $x\gt1$ and $1-\delta\lt x\lt1+\delta$. I would take $x=1+(\delta/4)$
That means $$|f(x)-1|=|x|=|1+\frac{\delta}{4}|=1+\frac{\delta}{4}\gt\frac{1}{2}$$ (since $\delta\gt0$)
That completes our proof. We can conclude now that $\lim_{x\to1}f(x)\ne1$
c) Prove that $\lim_{x\to1}f(x)\ne1.5$
Let $\epsilon=0.5$
Again, we need to show that for each $\delta\gt0$ there exists a value of $x$ such that $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1.5|\ge0.5$$
- For $x\gt1$, $f(x)=x+1$, meaning that $|f(x)-1.5|=|x+1-1.5|=|x-0.5|$
- Since $0\lt|x-1|\lt\delta$, that means $-\delta\lt x-1\lt\delta$, and $1-\delta\lt x\lt1+\delta$
Now we need to take a value of $x$ so that $x\gt1$ and $1-\delta\lt x\lt1+\delta$. I would take $x=1+(\delta/4)$
That means $$|f(x)-1.5|=|x-0.5|=|1+\frac{\delta}{4}-0.5|=|0.5+\frac{\delta}{4}|=0.5+\frac{\delta}{4}\gt0.5$$ (since $\delta\gt0$)
That completes our proof. We can conclude now that $\lim_{x\to1}f(x)\ne1.5$
