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So now that I have moved on to differentiation in multiple dimensions and a more general treatment, I recognize a derivative (let us treat the 1-d case for now, it is easy to generalize in more than 1 with norms and such) as an operator $D$ that takes $f \to f'$and $f'$ although not (necessarily) a linear function itself, takes points in the domain of $f$ and sends them to scalars $f'(x)$ which we say is the derivative. The mapping $g(h) = f'(x)\cdot h$ then satisfies:

$$f(x+h) = f(x) + f'(x) \cdot h + \epsilon_x(h)$$ where the error term $\epsilon_x(h)$ satisfies: $$ \frac{\epsilon_x(h)}{h} \to 0 $$ as $h \to 0$. Now, it is easy to see why the error term behaves this way by the limit definition of the derivative. My question here however, is why (intuitionally) does the quotient of the error term (rather than the error term itself) need to approach $0$? The more I think about it, the more I realize that the error term approaching $0$ itself is a trivial consequence of continuity, i.e the fact that $\lim_{h \to 0} f(x + h) = f(x) = \lim_{h\to 0}(f'(x)\cdot h + f(x))$. I guess my question is, why does the error (quotient) approaching $0$ give us a "good enough" or satisfactory linear approximation of $f$?

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In my answer linked below, I show that the tangent is the best linear approximation to a function at a point.

My answer is for functions of one variable, but I think it would be not too hard to generalize it to more dimensions.

How is the derivative truly, literally the "best linear approximation" near a point?

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  • $\begingroup$ Thank you, I was looking for something like this. $\endgroup$ – rubikscube09 Jan 11 '18 at 23:42
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If that weren't the case, then the "error" term would be asymptotically of at least the same order as the derivative term itself, which would mean that a better, still linear in $h$, approximation exists.

Another way to think about it is to consider the mean value theorem: $f(x+h)-f(x)=hf'(\xi)$ for some $\xi$ between $x$ and $x+h$ (depending on both points in general). What is $hf'(\xi)-hf'(x)$? Surely it goes to zero faster than $h$ if $f'$ is even just continuous. And if $f'$ is actually differentiable itself, then you get a better estimate...and this is the beginning of the line of thinking that gets you to Taylor's theorem.

(I realize that the latter point is deceptive in the case of a function differentiable at, say, just one point, but such pathologies are not generally good for building the foundation of intuition. Instead they are good for chiseling away the slightly incorrect edges of our initial intuition.)

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