Understanding Limits of Differentiability

[MarkFL]

This means that the two-sided limit using first principles must exist, but we need to bear in mind that $f(a)$ must also be defined.

edit: Actually, this still is not sufficient...consider:

$$f(x)=\begin{cases}-1 & x<0\\ 0 & x=0 \\ 1 & 0<x \\ \end{cases}$$

So, we should say $f$ must be continuous at $x=a$, rather than merely defined.

 

[Jameson]

Consider the function:

$$f(x)=\frac{x}{|x|}$$

It has a jump discontinuity at $x=0$, however, we find:

$$\lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=0=\lim_{h\to0^{+}}\frac{f(x+h)-f(x)}{h}$$

Is this the kind of thing you are after?

This is exactly the kind of example I was thinking of but couldn't come up with. So it seems that the existence of the limit that corresponds to the derivative of some function $f$ at point $a$ isn't sufficient enough to claim that $f(a)$ exists. So once again we have that differentiability implies continuity and that the left and right-hand limits that define $f'(a)$ exist. To go the other way and show differentiability, you need both of these though.

Any objections to the above claims?

 

[MarkFL]

This means that the two-sided limit using first principles must exist, but we need to bear in mind that $f(a)$ must also be defined.

edit: Actually, this still is not sufficient...consider:

$$f(x)=\begin{cases}-1 & x<0\\ 0 & x=0 \\ 1 & 0<x \\ \end{cases}$$

So, we should say $f$ must be continuous at $x=a$, rather than merely defined.

I made the above point about continuity in haste, the function is continuous, but we have:

$$\lim_{x\to0^{-}}f(0)\ne f(0)$$

$$\lim_{x\to0^{+}}f(0)\ne f(0)$$

$$\lim_{x\to0^{-}}f(0)\ne \lim_{x\to0^{+}}f(0)$$

So, what Jameson has posted above is fine with me as long as by continuity we mean:

$$\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=f(a)$$

 

[Reckoner]

Consider the function:

$$f(x)=\frac{x}{|x|}$$

It has a jump discontinuity at $x=0$, however, we find:

$$\lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=0=\lim_{h\to0^{+}}\frac{f(x+h)-f(x)}{h}$$

Is this the kind of thing you are after?

I'm not sure I understand. This limit only makes sense when $x \neq 0,$ right? Because $f(h) - f(0)$ is undefined.

Even with the other function you gave in your later post,
\[
f(x)=
\begin{cases}
-1, & x<0\\
0, & x=0 \\
1, & 0<x \\
\end{cases}
\]
we have, for $x = 0,$
\begin{align*}
\lim_{h\to0^-}\frac{f(x + h) - f(x)}h &= \lim_{h\to0^-}\frac{f(h) - f(0)}h\\
&= \lim_{h\to0^-}\frac{-1 - 0}h\\
&= \lim_{h\to0^-}-\frac1h = \infty
\end{align*}
and, similarly for the right-handed limit,
\begin{align*}
\lim_{h\to0^+}\frac{f(x + h) - f(x)}h &= \lim_{h\to0^+}\frac{f(h) - f(0)}h\\
&= \lim_{h\to0^+}\frac{1 - 0}h\\
&= \lim_{h\to0^+}\frac1h = \infty.
\end{align*}

These limits do not exist, which makes sense since $f$ is not differentiable at $0.$ Am I missing something?

 

[MarkFL]

Yep, that's what I get for relying too heavily on the computer:

d/dx(x/|x|) - Wolfram|Alpha

By hand, I get:

$$\frac{d}{dx}\left(\frac{x}{|x|} \right)=\frac{0}{|x|}$$

 

[Reckoner]

Can someone explain to me the above? Like how did f(h) become -1?

For all $x < 0,$ $f(x) = -1.$ Since we are taking the left-handed limit ($h$ is approaching 0 from the left), we are dealing with $h < 0.$ Similarly, for the right-handed limit we are looking at where $h > 0$ so that $f(h) = 1.$

 

[Deveno]

One small observation:

The quantity:

[math]\frac{(x+h) - f(x)}{h}[/math]

for [math]f(x) = \frac{x}{|x|}[/math]

has NO MEANING at the point [math]x = 0[/math] because f is UNDEFINED there.

In vague analytical terms, when we take the limit:

[math]\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}[/math]

we require three things:

1) f(a) is defined
2) f(x) is defined for some (possibly very small) open interval (a-|h|,a+|h|)
3) this limit exists

If one looks closely one sees that (1) and (2) together "almost" imply continuity, the limit condition (3) means we don't have some strange "skipping/jumping" behavior near the point a (in fact it says something STRONGER, it is a "smoothness" requirement)

The derivative f'(a) is a certain limit involving expressions with f. I want to point out that f'(a) is a (real) NUMBER, the FUNCTION f' (which has the value f'(a) at the point a, if it exists) is another matter entirely. For the FUNCTION to exist, we require differentiability at many,many points, and not all functions qualify (for example, the function f(x) = |x| has a "corner" at x = 0, and there exist extremely "jagged" functions, which are continuous everywhere but have no derivative anywhere!).

SO differentiabilty of a function (everywhere) is a rather SPECIAL quality, just as continuity is a special quality. In the grand scheme of things, MOST functions are not this special. In a way, calculus misleads us into thinking most functions SHOULD be continuous, or differentiable (because most of the ones we think of trying first, just so happen to qualify). The sad truth is this: the "bizarre" functions are the vast majority, the "nice" functions are a tiny sliver of these, which are the ones we concentrate on, just BECAUSE they are so EASY to work with.