Understanding Limits of Differentiability

[Dethrone]

I know this is probably a dumb question, but I have a question regarding this. My textbook says the following: "A function f is differentiable at a if f'(a) exists."

It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0⁺ and h -> 0^-. Finally concluding that it is not differentiable as the limits are different. Done.

My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.

EDIT:
Sorry if this is confusing, and if you want me to clarify, I can provide a hypothetical situation.

Spoiler

Ok, time to clarify what I'm trying to convey with a hypothetical situation (not real)

We are asked to see if f(x) = |x| is differentiable at x = 0. Let's PRETEND that the limit exists at f'(a). That when you took the limit from the positive and negative side of it, it resulted in 5. However, it is actually undefined at 5, but the limit as you approach f'(a) is 5. By using the same logic as above, the logic that the textbook example used, we are to assume that YES, It is differentiable, BECAUSE the limit exists! However, that is actually NOT the case, since the by the definition, f'(a) MUST exists. But in this example, YES, the limit exists, but f'(a) doesn't exist as it is undefined at a.

That being said, would additional steps be taken/required, in order to prove if something is or is not differentiable. As we see in my hypothetical situation, just determining the limit was not enough.

 

[Nono713]

Okay, this is hurting my brain so I'll conservatively assert that:

1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.

2. If the two limits are different, then for most well-behaved functions the derivative does not exist at that point, however there are may exist crazy analytical functions which are continuous while their derivative isn't, and vice versa, so it may not be a sufficient condition to show the existence of $f'(a)$.​

 

[Jameson]

This is a really good question.

[Rest of post deleted due to a mistake. Updated thought can be found in my later post]

 

[caffeinemachine]

Hey Jameson! :) I think what you said is not entirely correct.

This is a really good question.

In order for a function of one variable, $f(x)$, to be differentiable at a point $a$, then it must be be continuous at $a$ and the following limit must exist ...

I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.

2) Just knowing that the above limit exists is not enough to say that $f(x)$ is differentiable at $a$. ...

I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.

Please correct me if I am wrong or if I have misinterpretted your response.

 

[caffeinemachine]

I know this is probably a dumb question

This is not a dumb question at all. :)

My textbook says the following: "A function f is differentiable at a if f'(a) exists."
It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0⁺ and h -> 0^-. Finally concluding that it is not differentiable as the limits are different. Done.

My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.

I think you are making mistakes here. We write $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ whether the limit exists or doesn't exist. When the limit doesn't exist, $f'(a)$ is not defined. When the limit $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists, then $f'(a)$ is same as this limit. So $f'(a)$ is a real number and certainly it exists.

Note that when $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists then this doesn't say that $f'(a)$ has a limit. This would have no meaning. It says that $\frac{f(x)-f(a)}{x-a}$ has a limit as $x$ approaches $a$. For a short hand notation, and for furthering the theory, we denote this limit, whether it exists or not, as $f'(a)$.

I have not read the content in Spoiler box thoroughly yet. Tell me if you have further doubts or you need to discuss.

EDIT: I have implicitly assumed that $f$ is a real function having an interval $I$ in its domain such that $a\in I$.

 

[Nono713]

He is taking the limit of the derivative, not the original function itself.

 

[Jameson]

Hey Jameson! :) I think what you said is not entirely correct.

Yep, I think I need to edit a couple of things.

I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.

Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.

I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.

Please correct me if I am wrong or if I have misinterpreted your response.

About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course.

Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.

I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true. :)

 

[Jameson]

He is taking the limit of the derivative, not the original function itself.

Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.

 

[caffeinemachine]

He is taking the limit of the derivative, not the original function itself.

Umm.. now I am totally confused. :) If you understand his doubt then can you please frame it nicely and post it along with your solution?

 

[Nono713]

Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.

That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)

At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here :p haven't done any formal calculus in a while.

 

[Dethrone]

I haven't read all your responses yet, but it is the limit of the derivatvive. I don't knwo anymore haha, I'm too confused.

The textbook does this:

f'(x) = lim h-> 0 (|0 + h| - |0|)/ h
Then split in two parts to compute the left and right limits separetely.

 

[Jameson]

I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of the derivative". Your example seems to confirm this.

 

[Dethrone]

Yep, I think I need to edit a couple of things.
Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.
About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course.

Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.

I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true. :)

It's quite strange actually. I always do see functions written in teh form "f(x)" rather than just "f"... My textbook uses "f", so now I do that too. But my previous math teachers use "f(x)"

- - - Updated - - -

I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of a derivative". Your example seems to confirm this.

I'm confused. OH. My example was just finding the derivative of a function right, by taking its limit right? I got confused because I don't usually see the word "limit" when I take derivatives using power rule, chain, etc, and I got mixed up thinking it was the limit of a derivative.

And I haven't finished reading the responses yet from the previous page.

 

[Dethrone]

That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)

At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here :p haven't done any formal calculus in a while.

This is precisely what I meant with my question; I don't know if that is the limit definition of derivatives, or whatnot. Now if only my original post was as concise as how you explained it (Envy)

EDIT: second thought, brain hurts, not sure if that's how i meant it but:
It's just that I'm not convinced that f'(a) exists by just taking its limit from both sides. And it must exist for it to be differentiable at that point.

This:

1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.

I think the notation is confusing me; the limit definition of derivatives has both "f'(x)" and "lim" so it makes me want to think that it's taking the LIMIT of the DERIVATIVE, but now that I think about it, it's simply just taking the derivative.

 

[Dethrone]

This leads me to the following question:

When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.

Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?

For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0⁺ |x|= |h|/h = 1
f'(x)=lim as h ->0^- |x|= |h|/h = -1

Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0⁺ x / |x|
= x / x = 1
limit as h >0^- x / |x|
= x / -x = -1

Same answer.

 

[Nono713]

This leads me to the following question:

When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.

Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?

For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0⁺ |x|= |h|/h = 1
f'(x)=lim as h ->0^- |x|= |h|/h = -1

Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0⁺ x / |x|
= x / x = 1
limit as h >0^- x / |x|
= x / -x = -1

Same answer.

If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all) :)

I am not sure what you mean though - what would be the purpose of calculating such a limit?

The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because calculating derivatives from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic. The idea is that the limit definition gives the most "basic" definition of what a derivative is, which is important in mathematics (where something without a proper definition is useless).

 

[Dethrone]

If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all) :)

I am not sure what you mean though - what would be the purpose of calculating such a limit?

The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because deriving from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic.

Do you also mean to imply that finding the limit here is redundant?

Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0⁺ x / |x|
= x / x = 1
limit as h >0^- x / |x|
= x / -x = -1

So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0? (Instead of doing all the two side limit hassle that was only necessary when utilizing the limit definition of a derivative)

 

[Nono713]

Do you also mean to imply that finding the limit here is redundant?
So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0?

Well the same holds true for the function itself. Before differentiating your function you should rigorously show that it is in fact continuous at the locations you want to differentiate it at, otherwise the function is not differentiable there and you cannot even bring up the concept of derivative (it is meaningless). In this case, our function $f(x) = |x| / x$ has a discontinuity at $x = 0$ (you can rigorously show that by taking the left and right limits of the function around that point and demonstrating that $\lim_{x \to 0} f(x)$ does not exist as they are not the same ($-1$ and $+1$ respectively).

Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.

You need to be very careful when differentiating. Just because you get a nice expression for your derivative everywhere except at, say, $x = 0$, does not mean that expression is valid for $x = 0$! This is why it's important to keep track of the domain of each function you are working with. If you find that your derivative is valid for all $x \neq 0$, then that's it - end of story. It is meaningless to plug $x = 0$ into it, the results are undefined because the derivative doesn't apply at that point.

Also, limits don't have much to do with derivatives in the grand scheme of things, but they provide a useful framework on top of which to build the concept of derivative and differentiability (as well as a lot of other important stuff). Don't associate them with limits too much.

Here the "left and right limits" are used to show that the two-sided limit exists, and additionally show continuity. The limit expression of the derivative is different and has a geometrical interpretation, which is basically "take two points on a curve and draw a line between them, as the points get closer and closer together that line starts to approximate a tangent to the curve, which happens to also give the rate of change of the height of the curve with horizontal distance and lots of other useful things".

 

[Dethrone]

Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.

Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .

 

[Jameson]

You already showed that the left and right hand limits at $x=0$ are 1 and -1, which is all you need to do. That shows that the limit doesn't exist thus $g$ is not differentiable at $x=0$. Once again, differentiability implies continuity but continuity does not imply differentiability.

 

[Dethrone]

So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontious at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.

Is there anythign wrong with the procedure above?

 

[Nono713]

Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .

You take the definition of "differentiable":

A function $f(x)$ on $\mathbb{R}$ is differentiable at $x = a$ if and only if $f'(a)$ exists.

From that definition you can choose different approaches depending on your function. For instance, take $f(x) = |x|$. Then, assume $x < 0$, where the function is (probably) differentiable - we'll prove it is shortly. Now apply the definition of the derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

And we get:

$$f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h}$$

Now because $x < 0$, $|x| = -x$, and we can rewrite this as:

$$f'(x) = \lim_{h \to 0} \frac{|x + h| + x}{h}$$

Now assume $x + h > 0$. But this means that $h > -x$, and since $x$ is negative then $-x$ is positive so $h$ is greater than some non-zero value.. which is a contradiction since we assumed that $h \to 0$. Therefore $x + h < 0$, so $|x + h| = - (x + h) = -x - h$ and:

$$f'(x) = \lim_{h \to 0} \frac{-x - h + x}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 ~ ~ ~ \text{for} ~ ~ x < 0$$

A similar reasoning for $x > 0$ leads to:

$$f'(x) = +1 ~ ~ ~ \text{for} ~ ~ x > 0$$

So we have:

$$f'(x) = \begin{cases} +1 ~ ~ &\text{for} ~ ~ x > 0 \\ -1 ~ ~ &\text{for} ~ ~ x < 0 \end{cases}$$

What now? Well now the left and right limits around $x = 0$ are easy (note we are never actually touching zero, $f'(x)$ has not been defined at that point) and we clearly see that:

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

And we may conclude that:

$$\lim_{x \to 0} f'(x) ~ ~ ~ \text{does not exist} ~ ~ ~ \implies ~ ~ ~ f'(0) ~ ~ ~ \text{does not exist}$$

And therefore by the definition of differentiability, $f(x)$ is not differentiable at $x = 0$.

That was the semi-rigorous, long-winded way. In practice you use things like the chain rule and various limit/derivative theorems (or even plain observation) to help you simplify the function instead of starting from scratch each time.

For instance, using the result above, you now know that $f(x) = |x|$ is not differentiable at $x = 0$. From that point on you can easily show that $f(x) = |x - a|$ is not differentiable at $x = a$, for all $a \in \mathbb{R}$, and you can now use that fact to find other, more complicated derivatives involving absolute values, using the chain rule, product rule, and so on.

 

[Jameson]

So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontinuous at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.

Is there anything wrong with the procedure above?

I hope someone who is more versed in analysis stops by to comment on this, but I will add what I can.

1) Yes, this is a good start. There are other things to consider as well, such a bend, cusp or vertical tangent - but your step #2 will sort those out anyway. You'll develop an eye for noticing these things quickly.

2) It's always a good idea to be able to use the limit definition of a derivative (left and right hand limits as you said) but of course in practice you have quicker ways of calculating derivatives, as you know. You can probably tell from the problem if the goal is to quickly calculate the derivative and use that as part of a greater problem, or if the differentiability itself is the main focus of the problem. Anyway, this step is ok.

3) If the limit doesn't exist, it doesn't automatically follow that the function is not continuous at that point. $f(x)=|x|$ is a great example of this. It is continuous at $x=0$ but not differentiable.

The only thing that I am not certain of is whether step 1 is implied by step 2. Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.

Put another way, if we have a function of one real variable, $f$, can the limit following limit exist if the point, $a$ is not in the domain of $f$?

\(\displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\)

I think not because $f(a)$ is not defined. So that leads me to say that step 2 implies step 1.

 

[Dethrone]

Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.

This problem was what compelled me to start this thread in the first place. It occurred to me "what if the limit definition of a derivative results in an answer, but that point was actually discontinuous". I'm also interested if there such as a function as well, as it will clear my problems haha. I agree with you on the premise that step 2 implies step 1.

Thanks guys for the help!

 

[MarkFL]

Consider the function:

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

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

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

Is this the kind of thing you are after?

 

[Dethrone]

Consider the function:

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

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

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

Is this the kind of thing you are after?

Precisely, so for this example, just saying that it's continuous because the limit exists doesn't suffice, right? Since the point is undefined at that point.

Wait.

Isn't the \(\displaystyle \lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=1?\) and the other -1?

 

[MarkFL]

...
Wait.

Isn't the \(\displaystyle \lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=1?\) and the other -1?

No you are thinking of the function itself, not its derivative, consider the piecewise definition:

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

So, we see then that:

\(\displaystyle f'(x)=\begin{cases}0 & x<0\\ 0 & 0<x \\ \end{cases}\)

If you use the quotient rule to differentiate, you find $f'(x)=0$, but you then have to look (as I believe Bacterius stated) at the function's domain, which we should do first, and so we would then know that any results are meaningless at $x=0$.

 

[Dethrone]

No you are thinking of the function itself, not its derivative, consider the piecewise definition:

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

So, we see then that:

\(\displaystyle f'(x)=\begin{cases}0 & x<0\\ 0 & 0<x \\ \end{cases}\)

If you use the quotient rule to differentiate, you find $f'(x)=0$, but you then have to look (as I believe Bacterius stated) at the function's domain, which we should do first, and so we would then know that any results are meaningless at $x=0$.

Oh I see, by applying this:
"A function f(x) on R is differentiable at x=a if and only if f′(a) exists."

So does that imply that although the limit from both sides exist, x is not differentiable at 0?

 

[MarkFL]

I took this from Wikipedia, as it basically states what my understanding of the matter is:

If $f$ is differentiable at a point $x_0$, then $f$ must also be continuous at $x_0$. In particular, any differentiable function must be continuous at every point in its domain. The converse does not hold: a continuous function need not be differentiable. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

 

[Dethrone]

Thanks, i get it now.

Now for this: "function f(x) on R is differentiable at x=a if and only if f′(a) exists."
When it refers to f'(a), do they mean that it f'(a) exists when you take its limit and that the point itself can be undefined , or do they mean that the point f'(a) must exist on the graph and must be defined?

 

[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:

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

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

 

[Jameson]

Consider the function:

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

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

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

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:

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

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:

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

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

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

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

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

 

[Reckoner]

Consider the function:

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

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

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

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:

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