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:

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

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.