Answer
The limit of a function $f(x)$ as $x\to c$ may fail to exist if:
1) The values of $f(x)$ "jumps" at $x=c$
2) $f(x)$ approaches $\infty$ as $x\to c$
3) $f(x)$ oscillates as $x\to c$
Work Step by Step
The limit of a function $f(x)$ as $x\to c$ may fail to exist if:
1) The values of $f(x)$ "jumps" at $x=c$
Consider this function: $f(x)=0$ for $x\le0$ and $f(x)=1$ for $x\gt0$
As $x$ approaches $0$ from the left, $f(x)$ approaches $0$, but as $x$ approaches $0$ from the right, $f(x)$ approaches $1$. That means $f(x)$ does not approach any single value as $x\to0$, so its limit does not exist.
2) $f(x)$ approaches $\infty$ as $x\to c$
Approaching infinity means the limit does not exist, since it does not reach for any concrete value.
Consider the limit: $\lim_{x\to1}\frac{1}{x-1}$
As $x\to1$, $(x-1)$ approaches $0$, so $1/(x-1)$ will get infinitely large, or approach infinity. The limit fails to exist here.
3) $f(x)$ oscillates as $x\to c$
Consider this limit: $\lim_{x\to0}\sin\frac{1}{x}$
A drawing of the graph of the function $y=\sin(1/x)$ would reveal the mad oscillation of $f(x)$ around $x=0$. $f(x)$ does not approach any single value, so the limit fails to exist.
