Show that $f(x)$ is continuous at point $c$ I have a function $f(x) = 0$ on $[a,b]$ except for $c \in [a,b]$ where $f(c) = 1$.  I am asked to show $f(x)$ is continuous at point $c$.
Thus I have for the definition of continuity:
if $\;|x-c|<\delta\;$ then $\;|f(x)-f(c)|<\varepsilon\;,$
if $\;|x-c|<\delta\;$ then $\;|1|<\varepsilon\;.$
However given that the definition has to hold for all $\varepsilon>0$ then I'm not sure how this applies given that $\varepsilon>1$ by the above and thus there exist some values for which the definition does not hold.

Recall the definition of continuity: $f$ is continuous at $c$ if for all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x$, if $|x-a| < \delta$, then $|f(x) - f(c)| < \epsilon$.
I'm emphasizing the quantifiers.  When you negate a statement, the quantifiers flip and the negation moves down the line.  So $f$ is not continuous at $c$ if there exists $\epsilon > 0$ such that for all $\delta > 0$, there exists $x$ such that $|x-a| < \delta$ but $|f(x) - f(c)| \geq \epsilon$.
If you draw a graph of $f$ and inspect each of these definitions, I think you'll see which one is satisfied, and how.
