Answer
See explanations.
Work Step by Step
a. Examine the graph and we see a gap at $x=3$. For a given $\delta\gt0$, $|x-3|\lt\delta$ which gives $3-\delta\lt x\lt3+\delta$. In the interval of $3-\delta\lt x\lt3$, the function $f(x)\gt4.8$ which gives $|f(x)-4|\geq0.8$, thus we can set $\epsilon=0.8$ or less so that $|f(x)-4|\geq\epsilon$ for any $\delta\gt0$ which means that $\lim_{x\to3}f(x)\ne4$
b. Similarly, use the interval of $3\lt x\lt3+\delta$, the function $f(x)\lt3$ which gives $|f(x)-4.8|\geq1.8$, thus we can set $\epsilon=1.8$ or less so that $|f(x)-4.8|\geq\epsilon$ for any $\delta\gt0$ which means that $\lim_{x\to3}f(x)\ne4.8$
c. Again, use the interval of $3-\delta\lt x\lt3$, the function $f(x)\gt4.8$ which gives $|f(x)-3|\geq1.8$, thus we can set $\epsilon=1.8$ or less so that $|f(x)-3|\geq\epsilon$ for any $\delta\gt0$ which means that $\lim_{x\to3}f(x)\ne3$
