Question about finding the limit at an undefined point.

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This may be braindead, but I'm trying!

If I have a function $f$ and that function is not defined at some x, then asking for the derivative of the function at $x$ makes no sense since there is no $f(x)$ at $x$.

But if I want to find a gradient for that function as close as possible to x, then how does that work? Isn't that the same as the derivative at x? It's like, I can do the same calculation but I have to disregard the result because I'm asking for something that doesn't exist.

For example, if $f(x)=\frac{1}{x−2}$, then $f(x)$ is not defined at $x=2$. So I can't find the derivative at that point since it doesn't exist, But the limit is 2. But the limit is the derivative, and the derivative doesn't exist! I'm confused.

I felt like I understood this but I woke up this morning with no idea. Last week I was happily finding the volume of cylindrical wedges, now I can't understand limits O_O

Please set me straight.

EDIT: I think my problem is the way I'm thinking about limits. It seems that there are two limits and I'm confusing them. The limit that $f(x)$ approaches and the limit that $x+h$ approaches. In the above example where $f(x)=\frac{1}{x-2}$, $2+h$ approaches $2$ and $f'(x)$ is undefined since the numerator contains a division by zero.
Is that my answer?

2nd EDIT: This is what I'm really asking: How do I find $lim_{x\to a}f'(x)$?

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6 Answers

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Note: $\delta q$ is just one variable, not two. (Just incase that is confusing.)

Hey there, these are all great answers. But I'd like to be a bit more precise about what's confusing you, if I can:

If I have a function f and that function is not defined at some x, then asking for the derivative of the function at x makes no sense since there is no f(x) at x.

I think you're way too used to the variables and not seeing the entire point: When we take a derivative of a function (let's call it $z(q)$), we're taking a limit that eventually simplifies:

$ \frac{d}{dq} z(q)=\lim_{\delta q \to 0} \frac{z(q+\delta q)-z(q)}{\delta q} $

Now, I'm sure you're aware of that and all, but it seems you've confused your terms. You have used the term "$x$" in two different ways without realizing it. Think about your statement rewritten:

If I have a function $z(q)$ and that function is not defined at some point, then asking for the derivative of the function at that point makes no sense since there is no $z(q)$ at that point.

Notice how I didn't confuse the variable of the function with the point? Now, even that statement isn't entirely accurate. The question becomes: What are you precisely thinking? Consider: We have $z(q)$ and, let's presume, its derivative: $z'(q)$. Now, are you saying: Why should we consider $z'(x)$ (a specific value) when $z'(q)$ (a specific function) is discontinuous at $x$ (a specific point)? That makes perfect sense. Do you see the difference and how this clarifies precisely what you're saying and gets rid of the confusion? I think you understand this matter really well, you've just confused yourself by using the same thing to denote very different things.

I can't find the derivative at that point since it doesn't exist, But the limit is 2. But the limit is the derivative, and the derivative doesn't exist! I'm confused.

You've got a precise example of what I'm saying. You're confusing the limit of the derivative at the point with the limit of the function. Let me illustrate: Your derivative of $z$ is a specific limit. But it is NOT the same limit that you take when you take the limit of $z'$ at a point that $z'$ is undefined at. More annoyingly stated: $z'(q)=\lim_{\delta q \to 0}\frac{z(q+\delta q)-z(q)}{\delta q}$ whereas (the second) is $\lim_{q \to y}z'(q)=\lim_{q \to y}\lim_{\delta q \to 0}\frac{z(q+\delta q)-z(q)}{\delta q}$ (where $y$ is your undefined point).

I'm sorry if anything I've said here is useless or redundant, I just hope this helps. This is my first answer on math, so if I'm made any horrendous faux pas, I'm sorry.

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Let's take a real-valued function defined for $x>0$, let's assume that $f$ is differentiable for $x>0$, and finally let's assume that $\lim_{x\to0+}f'(x)=a$. I claim that $f$ can be continuously extended to $x=0$, that $f$ is differentiable at $x=0$ and that $f'(0)=a$.

There is a $\delta_0>0$ with $|f'(x)|\leq M:=|a|+1$ for all $x\in]0,\delta_0[\ .$ Given an $\epsilon>0$ put $\delta:=\min\{\delta_0,\epsilon/M\}>0$. Then by the MVT of differential calculus we have $$|f(x)-f(y)|\leq\epsilon$$ for all $x$, $y$ with $0<x<y<\delta$. By Cauchy's theorem it follows that the limit $\lim_{x\to0+} f(x)=:f(0)$ exists, and now $f$ is continuous at $0$.

Using the MVT again (note that for the MVT the function only has to be continuous at the endpoints) we can say that for any $x>0$ there is a $\xi\in\ ]0, x[\ $ such that $${f(x)-f(0)\over x-0}=f'(\xi)\ ,$$ and by assumption on $f'$ it follows that $f'(0)=a$.

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This question is too general, but let me try: compute the quantity in question first (presumably it is computable on, say, some open set for which the singularity in question is a boundary point), and then take the limit of the quantity as the parameter approaches singularity. For example, consider the function $f(x) = 2x^2 - 1$ if $x > 2$. We aren't defining $f(x)$ for any other values of $x$. The derivative of $f$ is $2x$ where it is defined, and $\lim_{x\rightarrow 2^+}f'(x) = 4$.

Of course, in the bigger picture, the bigger question is whether a given map can be extended to a larger domain (in some meaningful way) than the one it was defined on originally.

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But the derivative is also the gradient of the tangent at $x$.

Derivative of $f(x)$ at $a$ i.e. $f'(a)$ gives the slope of tangent touching the curve $f(x)$ at $x=a$. This means that if $f(a)$ is not defined, the slope of tangent touching $f(x)$ at $x=a$ is also not defined, i.e. $f'(a)$ is not defined.

But if I want to find a gradient for that function as close as possible to x, then how does that work? Isn't that the same as the derivative at x? It's like, I can do the same calculation but I have to disregard the result because I'm asking for something that doesn't exist.

yes, derivative at $x=a$ is the gradient of $f(x)$ at $x=a$.

If you want the gradient of $f(x)$ as close as possible to $a$, that means you are talking about the limit

$\displaystyle\lim_{x\to a}f'(x)$ i.e.

$\displaystyle\lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)\cdots(1)$

Now, we know that $\displaystyle f(x)=\frac{x^2-9}{x-3}$ is not defined at $x=3$, but that $\displaystyle\lim_{x\to 3}f(x)$ exists. It happens because $(x-3)\not=0$ as $x\to 3$and thus by cancellation, we have $f(x)=x+3$.

Now, if such a provision exists in our $f'(a)$, we can talk about the existence of limit in $(1)$. Otherwise it does not exist.

The derivative is the limit. But the derivative is also the gradient of the tangent at $x$. I can see how a limit can exist if $f(x)$ doesn't exist, but I can't see how a gradient can exist when $f(x)$ doesn't.

As I have said above, a limit say $\lim_{x\to a}f(x)$ exists even if $f(a)$ does not exist $\textbf{only in case f(x) has some 'uncomfortable at x=a but cancellable otherwise' part that enables taking of limit}.$

Other than such a case, the answer to your question would be:

If $f(a)$ is not defined, $f'(a)$ and the gradient of $f(x)$ at $x=a$ is not defined.

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Well if $f(x)$ is undefined then you certainly can't define $f'(x)$ using the conventional definition. However, it may be possible to make sense of a derivative at $x$ when $f(x)$ is undefined by considering

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

But I'm not sure how much sense you will be able to make of this for practical purposes.

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If the function $f$ was defined at a neighborhood of $a$ and also was continuous at the point $a$, then:

If $\lim\limits_{x\to a} f'(x)$ exists, then it is the value of $f'$ at $a$.

That is a case where the derivate at the point is the value of the limit. I thought you wanted to do something like this, but is not the case .

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