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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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    I understand what you mean. I want to avoid using infinitesimals. But I can't think of$a$way around it.2012-01-06

3 Answers 3

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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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    @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.2012-01-07
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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.2012-01-06