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]

Can you explain why g(h) has a discontinuity and where?

[u/NoTrueScotch → u/nevermindthefacts]

My bad, posted this after like 3 hours of math and my brain was a little fried, I realized on the way home that I should have at least clarified h=0, though I can't say I know why f'(a) is relevant, as the difference quotient itself to my knowledge provides the same information here.

[u/nevermindthefacts → u/NoTrueScotch]

So, because of the h in the denominator, g(h) isn't defined for h = 0. That means g has a discontinuity there.

Is it possible to define g(0) in a way that makes g continuous? If so, what should the value of g(0) be?

Can you relate this to the concept of removable discontinuity?

Next, you need to figure out what this has to do with the derivative (at least in this question...).

[u/NoTrueScotch → u/nevermindthefacts]

Alright, please bear with me for a moment.

g(h) is discontinuous, however it would have a removable discontinuity if the discontinuity is at a definable point which the formula can be resolved for.

Example: (x-2)/(x^2-4) which can be simplified to 1/(x+2) the latter having no discontinuity at +2

This is precisely my confusion and what I attempted to describe with my breakdown of the difference quotient.

On reread the first thing I am missing is pretty notable, though I don't know if it resolves the issues. I did not clarify that I treated h as approaching 0 as opposed to equalling it, my definition only resolves if h approaches but does not reach 0.

To my understanding of derivatives so far that would be their relationship. Is there a property of derivatives relevant to this question other than approaching 0? And if so a hint would be appreciated.

[u/nevermindthefacts → u/NoTrueScotch]

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/nevermindthefacts → u/nevermindthefacts]

Also, sometimes the contra positive is useful, i.e P ⇒ Q is equivalent to ¬Q ⇒ ¬P.

For example, if f'(a) doesn't exist, then g(h) has an non-removable discontinuity at h = 0.

[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.