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How do you justify such limit equalities as these?

$\lim \limits_{x \to 0} f(x) = \lim \limits_{bx \to 0} f(bx)= \lim \limits_{(x - 7) \to 0} f(x-7)$

I am familiar with the $\epsilon-\delta$ definition of the limit, I just don't know what it is that makes these statements equivalent:

$$ \forall \epsilon \ \exists \delta> 0 \ \forall x: |x| < \delta \implies |f(x) - l| < \epsilon$$ $$ \forall \epsilon \ \exists \delta> 0 \ \forall x: |bx| < \delta \implies |f(bx) - l| < \epsilon$$ $$ \forall \epsilon \ \exists \delta> 0 \ \forall x: |x - 7| < \delta \implies |f(x - 7) - l| < \epsilon$$

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  • $\begingroup$ I'm not an expert, but I have a guess. It seems to me, the thing is that $bx$ and $x-7$ are surjections. Consider some statement (or rather predicate) $P$, let's say $\forall x: P(bx)$. Since $bx$ is a surjection, I may conclude, that if the statement was true for all $x$, it should be true for all $bx$, e.g $\forall bx: P(bx)$. Renaming $bx \rightarrow x$ we get again $\forall x: P(x)$. $\endgroup$ – guest Nov 1 '19 at 0:33
  • $\begingroup$ Your notation is non-standard. You should rather write $\lim_{x\to 0}f(bx)$ and $\lim_{x\to 7}f(x-7)$. The expression under limit notation is always of the form $\text{variable} \, \to\, \text{value}$ and not $\text{expression} \, \to\, \text{value} $. $\endgroup$ – Paramanand Singh Nov 1 '19 at 2:17
  • $\begingroup$ @ParamanandSingh I infer this poster is being introduced to limits, and part of the point of this problem is to learn to manipulate the variables in a basic $\varepsilon-\delta$ proof. To justify what you've written, you also need to know that $\lim_{x \to c} f(g(x)) = f(g(c))$ when $f$ and $g$ are continuous. That's true, of course, but I'm guessing the poster hasn't gotten there yet. $\endgroup$ – Robert Shore Nov 1 '19 at 2:29
  • $\begingroup$ @RobertShore: yeah I also think so. That's why there is so much hue and cry over putting context into questions. But still few askers pay heed to it. $\endgroup$ – Paramanand Singh Nov 1 '19 at 2:31
  • $\begingroup$ This one certainly made a good faith effort to do so. $\endgroup$ – Robert Shore Nov 1 '19 at 2:32
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For the first equivalence, substitute $u_1=bx$. For the second equivalence, substitute $u_2=x-7$. Then you're just saying:

$$\lim_{x \to 0} f(x) = \lim_{u_1 \to 0} f(u_1) = \lim_{u_2 \to 0} f(u_2).$$

Make these substitutions in an $\varepsilon-\delta$ proof and you'll see they flow right through.

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  • $\begingroup$ Your statement mentioning the equality of limits is vacuosly true as the variable symbol is dummy. The limit is for a function $f$ at a point $c$ $\endgroup$ – Paramanand Singh Nov 1 '19 at 2:19
  • $\begingroup$ @ParamanandSingh That's precisely the point. $\endgroup$ – Robert Shore Nov 1 '19 at 2:25
  • $\begingroup$ OK got it. +1 there. $\endgroup$ – Paramanand Singh Nov 1 '19 at 2:31

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