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I've been working on understanding limits thoroughly, so I'm rewriting how I understand the chain rule. Please help me fill in my gaps in understanding.

$f$ is some function. Then

$f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Now I might want to evaluate something like

$$\left(f(g(x))\right)' = \lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$$

Evaluating this is tricky, so we need a way to do it.

$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h}$)

$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)})\cdot\lim\limits_{h \to 0}(\frac{g(x+h)-g(x)}{h})$

if $k=g(x+h)-g(x)$, then

$g(x+h)=k+g(x)$, so

$\left(f(g(x))\right)' = (\lim\limits_{h \to 0}\frac{f(g(x)+k)-f(g(x))}{k})\cdot(\lim\limits_{h \to 0}\frac{g(x+h)-g(x)}{h})$

and, assuming I can go ahead and just change the limit variable on the left term, then

$=(\lim\limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}) \cdot (\lim\limits_{h \to 0}\frac{g(x+h)-g(x)}{h})$

$=f'(g(x))\cdot g'(x)$

Which is easier to figure out, and is also the chain rule.

Is that correct?

  • 0
    You switch from treating $k$ as a function to treating $k$ as a constant, which is unjustified.2012-01-06
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    Where is $k$ constant?2012-01-06
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    When you take the limit as $k$ goes to zero. In principle I suppose you *could* do that with a function, but it requires more bookkeeping. I'd say you simply shouldn't use $k$ at all, it just makes it easy to forget that $k = k(x,h) = g(x+h)-g(x)$.2012-01-06
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    The statement that $b\lim_{h\to0} (a/h) = \lim_{h\to0} (ba/h)$ assumes $b$ is constant. What you really need is $\lim_{h\to0} (ab) = (\lim_{h\to0}a)(\lim_{h\to0}b)$, which can be proved by $\varepsilon$-$\delta$ methods.2012-01-06
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    Did you assume b was constant? I never said whether it was or not. In the example above, I in fact require b not to be, since k plays the role of b later. Your limits examples is useful though.2012-01-06
  • 1
    There is an error in the equality after "(which I don't know how to prove yet)": you cannot switch the $h$, which is the variable over which you are taking the limit, to *outside* the limit.2012-01-06
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    Thanks for the awful reformatting of my post, btw.2012-01-06
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    You cannot rewrite the limit so cavalierly, because of the problems that arise when $k(h)$ takes the value $0$ even if $h$ is not equal to $0$. (The line just before "and, assuming I can just go ahead..." is not valid in general). That's the real difficulty in this attempted manipulation, which is what leads to the defining an auxiliary function like Michael Hardy and I do; it would also help you *a lot* if you made $k$'s dependence on $h$ *explicit* by writing $k(h)$, instead of trying to hide it. Also, you keep writing $f'(g(x))$ when you mean $(f(g(x)))'$.2012-01-06

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