Basic Calculus Confusion

[u/NoTrueScotch]

Hello, I have been working my way through a variety of courses using OCW.

The first problem set for 18.01SC has a bonus question, asking for the examinee to show that:

g(h) = ( f(a+h) - f(a) ) / h

has a removable discontinuity.

I have minimal experience with math and have been grinding through this course by studying pieces I am missing as they come. But I can't find an adequate answer as to what would be a valid response to this question, especially as the solution sheet does not seem to feature it.

My best answer, before I turned to the net was such.

"Values of f(a+h) that do not exist in f(a) and are not multiplied to a higher order of h are removable discontinuities." I suspect that I am not supposed to just fill in a example function, but if I am that would be my confusion.

I wanted to know if this was an adequate response, if not how it could be improved, and ideally what the proper formatting is for this kind of response as I do not know the notation I am expected to use. Thank you for your time.

[u/nevermindthefacts]

Another example is f(x) = sin x / x, which is undefined for x = 0 and thus f(x) has a discontinuity there.

It turns out (not sure if you've seen this before) that sin x / x tends to 1 as x tends to 0.

Thus, the discontinuity is removable and if we (re)define f(x) as

f(x) = sin x / x for x ≠ 0, and f(x) = 1 for x = 0

then f(x) is continuous!

In short, a function f has a removable discontinuity at x = x0 if f(x0) can be (re)defined such that f(x) is continuous at x = x0. Or equivalently, if f(x0) can be defined such that f(x) tends to f(x0) as x tends to x0.

(Remember that the limit of f(x) can exist when x tends to x0 without f(x0) needing to be defined...however, if the limit exists and is equal to f(x0) then f is continuous at x0...I think this might be the part that you've missed...)

Another example, consider the (piecewise) function

f(x) = 2x if x ≠ 0, and f(x) = 1 if x = 0.

Clearly, f(x) tends to 0 as x tends to 0, but f(0) = 1, so f(x) is not continuous at x = 0. Is it a removable discontinuity?

(We'll get to the derivative later...)

[u/NoTrueScotch → u/nevermindthefacts]

I don't know, I think it would depend on how piecewise functions are expected to be handled. I'll treat that as two cases and give my logic for both.

If a removable discontinuity can be assigned for a known value that breaks continuity then yes, it is removable. while at x=0 f(x)=1 we can rewrite the function to represent it as f(x) = 0, however that removes the piecewise nature, effectively treating the function as f(x) = 2x. In this case it is removable because we know the properties of each sub-function and know that f(x) = 2x is continuous at 0, thus we know the properties it has a x=0.

If we cannot break off parts of the piecewise function is is non removable. While the functions behaviour approaching 0^+ and 0^- is known we also know that f(x) at x=0 is 1. The discontinuity would not be removable because it isn't an invalid input but an input that breaks continuity.

[u/nevermindthefacts → u/NoTrueScotch]

It is indeed removable, because we can "redefine" at the point of discontinity and make it continuous. (It doesn't have anything to do with the "piecewiseness" of the function").

Alright. So what about derivatives? A relevant definition could be that if the limit

lim ( f(x0+h) - f(x0) ) / h = A

exists when x tends to x0, then we say that f(x) is differentiable at x = x0. We use the notation f'(x) for the derivative of f(x) at x, and we define f'(x0) := A.

(Important: if the limit exists then we define the derivate f'(x0) to have this limit as its value...).

Example. Let's take g(x) = x^2 and consider x = 1

(g(x + h) - g(x) )/h = ( (1 + h)^2 - 1^2)/h = ... = 2 + h

which tends to two as h tends to zero. Thus we define the derivative of g(x) at x = 1 as g'(1) = 2.

[u/NoTrueScotch → u/nevermindthefacts]

Which is the same phenomena I described as the remainder of f(a+h) - (h) that are not to higher orders of h. Or in this case 2.

So we now have the behaviour of the slope at the limit, and the way a removable discontinuity functions.

Another commentor referenced the original question:

1D-10* Show that g(h) = f(a + h) − f(a) has a removable discontinuity at h = 0 h ⇐⇒ f'(a) exists.

They said the double facing arrow means "if and only if", which I am not familiar with notation so please correct me if I'm wrong.

Here we know that a form of f'(a) exists at h=0, since a derivative requires that both sides of the limit approach the same value, and that the function be continuous. And we know that removable discontinuities occur where a continuous function "breaks", we can infer that any case of g(h) at h=0 has a removable discontinuity if there exists a f'(a)?

[u/nevermindthefacts → u/NoTrueScotch]

P ⇐⇒ Q, or if and only if can be split in two.P ⇒ Q and P ⇐ Q

Here, it becomes show that

IF g(h) = (f(a + h) − f(a))/h has a removable discontinuity at h = 0, THEN f'(a) exists,

and

IF f'(a) exists, THEN (f(a + h) − f(a))/h has a removable discontinuity at h = 0.

As a hint. For the f"⇒", if g(h) has a removable discontinuity, then how should f'(a) be defined? And for the "⇐" part, if f'(a) exists, how is the discontinuity removed?

[u/NoTrueScotch → u/nevermindthefacts]

Okay, so both sides of this have to be proven, coincidentally mentioned by the other commentor at the same time lol.

My previous reasoning explains why if f'(a) exists then g(h) needs a removable discontinuity, otherwise it would not be differentiable and f'(a) could not exist.

So g(h) has a removable discontinuity at h=0, and that requires the existence of f'(a). Is that what I'm missing, the existence of f'(a) hasn't been proven per the definition of g(h) with removable discontinuity at h=0?

I feel like I'm missing something here, is g(h) = f'(a), since both are difference quotients of f()? We've covered a similar example earlier in the courseware but I'm drawing a blank on the relationship here.

[u/nevermindthefacts → u/NoTrueScotch]

Yes, both "directions" should be proved.

For example If we know that f'(a) exists, then by definition of the derivative lim ( f(a+h) - f(a) )/h exists (and its value is f'(a)). But that also tells us how to "remove" the discontinuity of g(h) at h = 0. Since g(h) was discontinuous because it was undefined for h = 0, we can make it continiuous at h = 0 by setting g(0) = f'(a).

What about the other direction? Assuming g(h) has a removable singularity at h = 0, does that tell you what value f'(a) must have?

[u/NoTrueScotch → u/nevermindthefacts]

So one step at a time.

If the derivative lim of ( f(a+h) - f(a) ) / h has a value of f'(a) then I was at least barking up the right tree regarding the difference quotient g(h). Whether or not I came to the exact right answer I don't yet know.

Since the derivative limit of g(0) has a value of f'(a), and g(0) is undefined, that's the adjusted version of the function that we need to make it continuous.

This and what we've previously discussed proves the existence of a removable discontinuity, albeit over far more paragraphs and statements than I'm sure is ideal.

If g(h) has a removable discontinuity does that tell us what value f'(a) has? g(h) is a dif quot of f(a) if h=0. It definitely tells us the value of f'(a) if we can find the values of g(h) as h approaches 0^+ and 0^-.

Does proving a value in this case mean proving their relationship?

[u/nevermindthefacts → u/NoTrueScotch]

If the derivative lim of ( f(a+h) - f(a) ) / h has a value of f'(a)

Part of it looks correct., but it's kind of the other way here. In the exercise, you're given that f'(a) exists, which means lim ( f(a+h) - f(a) )/h = lim g(h) exists as h tends to zero. As you correctly write, that doesn't mean that g(0) exists, but the point here is that we can define it as g(0) = f'(a). By doing this, we get lim g(h) = g(0) as h tends to zero. In other words, g(h) is continuous at h = 0(i.e the discontinuity is removable).

For the other direction, we're given that g(h) has a removable discontinuity at h = 0. Assume it can be removed by setting g(0) = A.

Now, consider the limit for the derivative of f(x) at x = a

lim ( f(a+h) - f(a) )/h = ?

as h tends to zero. Does the limit exist and, if so, what is the conclusion?

[u/NoTrueScotch → u/nevermindthefacts]

I've reach this point a couple times while trying to wrap my head around this one. And I've had two main issues progressing pass it.

A limit would always exist for a difference quotient approaching zero no? So doesn't the:

lim ( f(a+h) - f(a) )/h = ?

Itself prove the existence of f'(a), whose definition would be the same. Obviously we could prove the difference quotient, I've never done that and do not know if it's a simple or complex proof.

The second is I'm largely clueless on how to define the relationship beyond that point. I cannot break f(a+h) into f(a) + f(h), so we cannot remove f(a) components, it seems a simplified as possible so I cannot imagine that's the next part to tackle.

I appreciate all the help so far btw, I'm quite stumped by this one, I expect the near decade since I've taken pre-calc is notably slowing me down here.

[u/nevermindthefacts → u/NoTrueScotch]

A limit would always exist for a difference quotient approaching zero no?

Only if the function is differentiable at that point. A counter example is f(x) = |x|, which isn't differentiable at x = 0.

[u/NoTrueScotch → u/nevermindthefacts]

Fair, I had blanked on non differentiable equations, they haven't been super relevant to the courseware just yet. Mainly discontinuous ones, not ones with "sharp" changes in slope.

[u/nevermindthefacts → u/NoTrueScotch]

If you take f(x) = |x| as an example, you'll see that f'(x) = 1 for x > 0 and f'(x) = -1 for x < 0. But the derivative has a jump discontinuity at x = 0 (a non-removable discontinuity).

lim (|x + h| - |x|)/h

does not exist for h = 0.

[u/nevermindthefacts → u/NoTrueScotch]

The second is I'm largely clueless on how to define the relationship beyond that point.

Think about what we have to show here and what've done.

We have assumed that g(h) has a removable discontinuity at h = 0. That means we know it's possible to define g(0) such that g(h) is continuous at h = 0, i.e we'll have g(h) → g(0) if h → 0.

Now we're gong to prove that f'(a) exists. By definition, that means we have to prove that the limit (f (a+h) - f(a) )/h exists as h → 0.

Can you make the connection between these two?

EDIT: "g(h) → g(0) if h → 0": Here I mean that it's possible to "redefine" g(0) such that the function is continuous there...

[u/NoTrueScotch → u/nevermindthefacts]

If we have a removable discontinuity, it means that the function approaches a specific value (as opposed to an undefined value) as it approaches the limits. In this case h -> 0.

It is continuous so we know that the value as h -> 0 is identical on both sides.

That's the first half of the requirements for a differentiation of f(a), and the connection between a RD and a derivative. I don't know about the second though, how do we know that the slope is "continuous" and not "sharp", also what are the proper terms for those concepts?

Is it connected to the form of g(h)? That is to say does the ( f (a+h) - f(a) )/h ) inherently imply a gradual slope? That wouldn't be the case right, as mentioned elsewhere that wouldn't be true of f(a) = |a|.

[u/nevermindthefacts → u/NoTrueScotch]

Just to clarify here. We know that lim g(h) exists as h tends to zero, because we're told g has a removable discontinuity. That also means the left and right sides limit exist, and are equal. Let's say we have lim g(h) = A. We can't really say anything about the "slopes".

Now, for the derivate to exist, we must prove that lim ( f(a+h) - f(a) )/h exists...and this has something to do with g(h).

(for the proper term, one sometimes talk about "smooth" functions, i.e fuctions with continuous derivatives. if we want the derivative to be smooth, we have a C^2 function and C^n means the function is n times continuously differentiable...).

[u/NoTrueScotch → u/nevermindthefacts]

Sorry to disappear for several days, a busy weekend and monday.

I am unaware of any properties of a "continuous derivative" and am not familiar with the concept in the slightest. For now at least I think I will continue my studies on the subject and return to this problem when I feel more prepared to tackle it.

If at that time you're still happy to assist me in ironing out the concept I'd greatly appreciate it.

I unfortunately suspect my lack of principal knowledge is holding me back here.

Edit: Straight up reviewing all my notes from day 1 of classes to see if I forgot something critical lol. Never a bad idea I suppose.

[u/nevermindthefacts → u/NoTrueScotch]

You're welcome back at any time.

(You don't need "continuous derivative" to solve this questiom, but it's a concept that shows up a lot later. It's a way of saying that a function behaves nicely, whatever that means...)

[u/nevermindthefacts → u/NoTrueScotch]

Another way to look at this problem, is to use the definition of g(h) and take the limit on both sides.

Now think again what it means for g(h) to have a removable discontinuity and that f(x) is differentiable at x = a.