Answer
To show that $\lim_{x\to3}f(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-3|\lt\delta\Rightarrow|f(x)-M|\ge\epsilon$$
Work Step by Step
To show that $\lim_{x\to3}f(x)\ne M$, choose a $\epsilon\gt0$, then prove that for each $\delta\gt0$, there exists a value of $x$ such that $$0\lt|x-3|\lt\delta\Rightarrow|f(x)-M|\ge\epsilon$$
a) $\lim_{x\to3}f(x)\ne 4$
Let $\epsilon=0.5$
- For $x\gt3$, we see from the graph that $f(x)\lt3$.
Therefore, $f(x)-4\lt3-4=-1$
Hence, $|f(x)-4|\gt1\gt0.5$
That completes our proof. We can conclude now that $\lim_{x\to3}f(x)\ne4$
b) Prove that $\lim_{x\to3}f(x)\ne4.8$
Let $\epsilon=0.5$
- For $x\gt3$, we see from the graph that $f(x)\lt3$.
Therefore, $f(x)-4.8\lt3-4.8=-1.8$
Hence, $|f(x)-4.8|\gt1.8\gt0.5$
That completes our proof. We can conclude now that $\lim_{x\to3}f(x)\ne4.8$
c) Prove that $\lim_{x\to3}f(x)\ne3$
Let $\epsilon=0.5$
- For $x\lt3$, we see from the graph that $f(x)\gt4.8$.
Therefore, $f(x)-3\gt4.8-3=1.8$
Hence, $|f(x)-3|\gt1.8\gt0.5$
That completes our proof. We can conclude now that $\lim_{x\to3}f(x)\ne3$
