# How do I evaluate a statement like $\lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)$?

I need to know how to do this to find out the derivative of a function as close as possible to an undefined point.

Assume that $f$ is not differentiable at $x=a$.

This might be the same as asking how to evaluate

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

Not sure.

Basically, what's the derivative of a function $f$ at the point closest to $x=a$ if $f$ is not defined at $a$? Note I'm not asking for the derivative of $f(a)$, since that is undefined, but instead the point closest to the undefined point, the endpoint I suppose.

What's the endpoint and what's the derivative of it?

I'm guessing I have to break this into a piecewise function but I'd rather not.

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It depends very much on the context. What is $f$? If you can find an expression for $f'(x)=\lim\limits_{h\to0}\frac{1}{h}(f(x+h)-f(x))$ for $x$ sufficiently close to $a$, then it is just a matter of evaluating the limit of the function $f'$ at $a$, provided these limits exist. – Jonas Meyer Jan 6 '12 at 19:39
if f is given function ,then in brackets statement is equivalence f',so it will be limit of f' at point x=a – dato datuashvili Jan 6 '12 at 19:42
There is no point different from $a$ and closest to $a$. – André Nicolas Jan 6 '12 at 21:32
Why not? I feel like I'm getting into infinitesimal territory and that's exactly what I'm trying to avoid. – Korgan Rivera Jan 6 '12 at 21:38
@Korgan Rivera: Undivisible infinitesimal theory is what you are getting into. I should have said there is no real number closest to $a$ and different from $a$. Suppose to the contrary there is one, say $b\ne a$. Look at $|b-a|$. It is a positive real number and has the property that there is no positive real number less than $|b-a|$. But that cannot be true, $|b-a|/2$ is less than $|b-a|$. So there cannot be a real number different from $a$ but closest to $a$. (There cannot be an element $b$ of an extension of the reals either, if division by $2$ is possible there.) – André Nicolas Jan 6 '12 at 22:24

Although you already have a couple of answers to your question I would like to add something which you might find clarifying. As other users said, in your case, you should first evaluate the inner limit (that is, if possible, obtain a general expression for $f'(x)$ in a neighbourhood of $a$), and then take the limit for $x \rightarrow a$.
What I wanted to point out is that, even if you assume that $f$ is differentiable at $x=a$ this is not enough to conclude that
$(1)$ $\lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)= \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)$.
Take for example $f(x) = \begin{cases} x^2\sin(\frac{1}{x}) \ \text{if} \ x \neq 0 \newline 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\ x=0\end{cases}$
You can check that $f$ is differentiable at $x=0$ (through the definition of derivative) but if you compute the derivative $f'(x)$ for $x \neq 0$ and take the limit $\lim_{x \rightarrow 0 }f'(x)$ the you will notice that such limit doesn't exist. In other words $f$ is differentiable at $x = 0$ (and in all of $\mathbb{R}$), but $f'$ isn't continuous at $x=0$. However if the limit $\lim_{x \rightarrow 0 }f'(x)$ did exist, you would be able to conclude that $f$ is differentiable at $x=0$, in other words it is possible to prove the following

Theorem if $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable in a neighbourhood of $a \in \mathbb{R}$, and $\lim_{x \rightarrow a }f'(x)$ exists and is finite, then $f$ is differentiable at $x=a$ and $\lim_{x \rightarrow a }f'(x)= f'(a)$. (Analogous versions hold if the limit isn't finite)

Concluding $(1)$ without any particular reason (such as the hypothesis of the theorem) is very tempting, but it is also a common mistake in calculus. Hope this helps.

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First evaluate the limit inside, that is, $\lim_{h\rightarrow 0}\frac{f(x + h) - f(x)}{h}$, and then the limit outside. If $f$ is differentiable at $x$, then the limit inside is just $f'(x)$. So then you're taking the limit $\lim_{x\rightarrow a}f'(x)$. For example, suppose $f(x) = 2x^2 - 1$. The derivative of $f$ is $f'(x) = 4x$, and $\lim_{x\rightarrow a}f'(x) = f'(a) = 4a$.

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The question was edited after your answer to say that you should assume $f'(a)$ does not exist. – Graphth Jan 6 '12 at 20:40
Yep, I've updated it again if you'd like to check it out. I figured what I'm really asking for is how to figure an endpoint and then find the derivative of it. – Korgan Rivera Jan 6 '12 at 21:21
In that case, see your previous question (math.stackexchange.com/questions/96990/…). By the way, you cannot hope for a straightforward answer to a question so general. The basic idea is this: if you know that arbitrarily close to the point in question the derivative DOES exist, then evaluate the derivative there, and then take the limit of the DERIVATIVE to the point in question. This is what people were suggesting (in a million different ways) on your previous question. – William Jan 7 '12 at 3:58
@WNY It required me to ask this question and receive some feedback for me to truly understand what I really needed to ask. I understand now that my question as it initially stood was deficient. Only now do I understand that I want to do as you've said: to find the derivative of the point that is arbitrarily close to the 'no-fly zone', where $x=a$. My next question is how do I find that point? Can I do this with my current 'tools'? I suspect not. – Korgan Rivera Jan 7 '12 at 6:57
@Korgan: How do you find which point? The one where the function is undefined? This should be derivable from the explicit form of your function (I presume you have the function given explicitly). For example, if $f(x) = \frac{x^2 + x - 1}{x - 1}$, then $f$ is not defined at $x = 1$ (a priori at least). If you want to find an point arbitrarily close to the "no fly" zone - well, that's the point of taking limits. – William Jan 7 '12 at 7:34

Start with an example. How about $f(x) = \frac{1}{x}$, which is defined for all $x \neq 0$. Then, the inner limit is $f'(x) = -\frac{1}{x^2}$, which is also defined for $x \neq 0$. Taking the limit, $\lim_{x \to 0} -\frac{1}{x^2}$, we get $-\infty$. You could now start with $g(x) = -\frac{1}{x}$ to find a final limit of $+\infty$. You could now start with $h(x) = \frac{1}{x^2}$ to end up with left hand and right hand limits that are different, so the limit does not exist at all, as a real number or even an extended real number.

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This is a useful example. – Korgan Rivera Jan 6 '12 at 21:15