If $H$ is the Heaviside function, prove, using the definition below, that $\lim \limits_{t \to 0}{H(t)}$ does not exist.


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 $$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$


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.


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


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?

  • $\begingroup$ Looks like you have everything you need. Show that definition cannot be satified with $\epsilon = \frac 12$ i.e. $\forall \delta >0, \exists t$ such that $|t|<\delta$ and $|H(t)-L|>\epsilon$ $\endgroup$ – Doug M Nov 23 '16 at 21:05
  • $\begingroup$ Mind explaining it without the upside-down letters? :) I'm not familiar with that notation. $\endgroup$ – Fine Man Nov 23 '16 at 21:06
  • $\begingroup$ Also, shouldn't $|H(t)-L|>\epsilon$ be $|H(t)-L|<\epsilon$? $\endgroup$ – Fine Man Nov 23 '16 at 21:09

$\forall$ means "for all" or "for any"

$\exists$ means "there exists"

For any $\delta > 0$ there exist a $t$ such that $|t|<\delta$ and $|H(t) - L|>\epsilon$

What are we doing.

If the limit exist. Then:

For any $\epsilon > 0$ there exist a $\delta>0$ such that when $|t|<\delta,|H(t) - L|<\epsilon$

And what does that mean. When $t$ is in the neighborhood of $0, H(t)$ is in a neighborhood of $L.$ The $\epsilon-\delta$ tell us the size of the neighborhoods.

But we need to show that the limit does not exist. There is a $t$ in the neighborhood 0, such that $H(t)$ is not in the neighborhood of $L$

How to do that? Fix the size of the target neighborhood of $H(t)$ small enough such that no matter how small of a neighborhood we find around $t$ we are sure to miss the target neighborhood around $H(t)$

  • $\begingroup$ I'm not quite understanding you. Can you be more concrete? $\endgroup$ – Fine Man Nov 23 '16 at 21:20
  • $\begingroup$ No matter how tight a neigborhood around $0$ we try to bind $t,$ there is a $t$ in the neighborhood such that $H(t) = 0$ there is also $t$ in that neighborhood such that $H(t) = 1.$ That means with $\epsilon = \frac 12$, for some $t, |t|<\delta, |H(t) - L|>\epsilon$ There is no $L$ that satisfies the definitions necessary for the limit to exist. $\endgroup$ – Doug M Nov 23 '16 at 21:25

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.


Main idea: Set $\epsilon = 0.1$ and see that, for any "candidate to limit" $L$, you will always find a point $X$ as close to $0$ as you want, such that $|H(X)-L| > \epsilon$

So, we've found an $\epsilon > 0$ such that, for any $\delta>0$, there exists $x$ such that $|x-0| < \delta$ but yet $|H(X)-L| > \epsilon$. But how to be precise about finding that $x$?

If our "candidate to limit" $L \in [0.9, 1.1]$, simply choose $x = - \frac{\delta}{2}$, so that $H(x)=0$, which is at a distance larger than $\epsilon$ from $L$, i.e. $|f(x)-L| > \epsilon$

I think you can easily guess which $x$ to pick if $L \notin [0.9, 1.1]$


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