Exercise
If $H$ is the Heaviside function, prove, using the definition below, that $\lim \limits_{t \to 0}{H(t)}$ does not exist.
Definition
Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write
$$\lim \limits_{x \to a}{f(x)} = L$$
if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that
$$\text{if } 0 < |x – a| < \delta \text{ then } |f(x) – L| < \epsilon$$
Hint
Use an indirect proof as follows. Suppose that the limit is $L$. Take $\epsilon = \frac{1}{2}$ in the definition of a limit and try to arrive at a contradiction.
Attempt
Let $\delta$ be any (preferably small) positive number.
$H(0 – \delta) = H(-\delta) = 0$
$H(0 + \delta) = H(\delta) = 1$
$H(0 – \delta) =^? H(0 + \delta) \implies 0 =^? 1 \implies 0 \neq 1 \implies H(0 – \delta) \neq H(0 + \delta)$
$\lim \limits_{t \to 0^-}{H(t)} \neq \lim \limits_{t \to 0^+}{H(t)} \implies \lim \limits_{t \to 0}{H(t)}$ does not exist
Request
I don't even know where to begin, even with the hint.
Can someone kickstart the proof for me?$^1$
$^1$ Update: I've come up with an attempt. Is it valid? It seems that I don't use the hint to my advantage; so if indeed my attempt is correct, what is the alternative proof using the hint?
Best Answer
Your attempt expresses the right idea but doesn't directly use the definition.
Here is one way to do it following the hint:
Suppose $\lim_{t\rightarrow 0}H(t)=L$. Then for every $\epsilon >0$, there exists $\delta >0$ such that $\left | H(t)-L \right |< \epsilon $ if $\left | t-0 \right |=\left | t \right |<\delta $. In particular it must work for $\epsilon =\frac{1}{2}$.
Take $t=\frac{\delta }{2}<\delta $ and $t^{'}=-\frac{\delta }{2}> -\delta $. Then we have:
$\left | H(t)-L \right |=\left | 1-L \right |< \frac{1}{2}$
Similarly, we have:
$\left | H(t^{'})-L \right |=\left | 0-L \right |=\left | L \right |< \frac{1}{2}$
Now, using the triangle inequality:
$1=\left | L-1+L \right |\leq \left | L-1 \right |+\left | L \right |< \frac{1}{2}+\frac{1}{2}=1 $
But wait! We have reached the following contradiction: $1< 1$
We have thus shown that $\lim \limits_{t \to 0}{H(t)}$ doesn't exist.