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I am reading Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms, and in it they provide the following theorem and proof. My issue is not with the theorem, but more so with the proof. Note that, in the theorem below, Equation 1.7.20 refers to the basic definition of the (multivariable) derivative, and $L$ is the linear transformation which defines the derivative. enter image description here

In particular, they seem to implicitly be distributing the limit in order to make the desired conclusion. That is, they seem to be using the theorem below (bullet point 1). enter image description here

But this theorem of course requires the hypothesis that both $f$ and $g$ have the given limit. Translating to our case in the proof of Prop. 1.7.11, use of Theorem 1.5.26 in the proof would require a priori knowledge that $$ \lim_{\mathbf{h}\to \mathbf{0}}\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})$$ exists (the second term in the limit of course exists, as justified). But how can we know this without resorting to the epsilon-delta definition of the limit? Indeed, that this limit exists (and is 0) is precisely what we are trying to prove. Thus, is this one of those instances where the authors intend for you to fill in the details, or is there really some way to deduce the result without resorting to epsilons and deltas (I imagine it's quite simple).

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  • $\begingroup$ Are we all supposed to know what equation 1.7.20 is? $\endgroup$ Nov 17, 2020 at 14:01
  • $\begingroup$ Indeed, my apologies - that is simply the basic statement of the limit definition of the (multivariable) derivative. $\endgroup$
    – EE18
    Nov 17, 2020 at 14:03

1 Answer 1

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No, it is not an argument by contradiction. If$$\lim_{h\to0}\frac{f(a+h)-f(a)-L(h)}{\|h\|}=0,$$then$$\lim_{h\to0}\bigl(f(a+h)-f(a)-L(h)\bigr)=\lim_{h\to0}\|h\|.\frac{f(a+h)-f(a)-L(h)}{\|h\|}=0$$and therefore, since $\lim_{h\to0}L(h)=0$ (since $L$ is linear and continuous), we have$$\lim_{h\to0}f(a+h)=\lim_{h\to0}\bigl(f(a+h)-f(a)\bigr)+f(a)=0+f(a)=f(a).$$

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  • $\begingroup$ Thanks very much for the answer! Unfortunately, I'm afraid I still don't follow (starting from the "we have" on the penultimate line). Why does $\lim_{h\to0}\bigl(f(a+h)-f(a)\bigr)$ necessarily exist (and in particular equal 0), as it seems has been assumed in the first equality (in the last line) in order to distribute the limit? I'm sure this follows immediately somehow from the hypothesis, but I can't seem to see it. Would you be able to justify it explicitly for me? $\endgroup$
    – EE18
    Nov 17, 2020 at 15:29
  • $\begingroup$ Ah! I might see it now. Is it that $\lim_{h\to0}\bigl(f(a+h)-f(a)-L(h)+L(h)\bigr)=\lim_{h\to0}\bigl(f(a+h)-f(a)-L(h)\bigr) + \lim_{h\to0}L(h)=0$ where we may split up the limit because we know that each term's limit exists, as well as what they equal, so we obtain $0=\lim_{h\to0}\bigl(f(a+h)-f(a)\bigr)$? $\endgroup$
    – EE18
    Nov 17, 2020 at 15:32
  • $\begingroup$ Yes, that is correct. $\endgroup$ Nov 17, 2020 at 16:46
  • $\begingroup$ Superb, thank you again! $\endgroup$
    – EE18
    Nov 17, 2020 at 16:54
  • $\begingroup$ I'm glad I could help. $\endgroup$ Nov 17, 2020 at 17:02

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