1

I'm really obsessed with using substitution in order to finding limits . When we can use that ? (What's the conditions for using substitution ?)

What's the rigorous proof which enables us to do that ? For example is that true if we say : $\lim_{x \to \infty } x - \lfloor x \rfloor $ doesn't exist so we conclude $\lim_{x \to 0 } 1/x - \lfloor 1/x \rfloor $ also doesn't exist ?

S.H.W
  • 4,379
  • 2
    Similar: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits, https://math.stackexchange.com/questions/1069642/finding-a-limit-using-change-of-variable-how-come-it-works – Hans Lundmark Feb 01 '18 at 18:46
  • 1
    "is that true if we say : $\lim_{x \to \infty } x - \lfloor x \rfloor $ doesn't exist so we conclude $\lim_{x \to 0 } 1/x - \lfloor 1/x \rfloor $ also doesn't exist ?" Let's be careful. You don't want $x\to 0,$ you want $x\to 0^+$ The general result you want in this setting is: Assume $f$ is defined in some interval $(a,\infty).$ Then $\lim_{x\to \infty} f(x) = L$ iff $\lim_{x\to 0^+} f(1/x) = L.$ – zhw. Feb 11 '18 at 16:30

2 Answers2

3

A standard theorem about limits, which you can probably find a proof of in a standard calculus text is: If $\lim_{x\to a}f(x)=b$ and $\lim_{x\to b}g(x)=c$ then $\lim_{x\to a}g(f(x))=c$.

Proof: Let $\epsilon>0$. Since $\lim_{x\to a}f(x)=b$ and $\lim_{x\to b}g(x)=c$, there are $\delta_1,\delta_2>0$ such that $|x-a|<\delta_1$ implies $|f(x)-g|<\delta_2$, and $|x-b|<\delta_2$ implies $|g(x)-c|<\epsilon$.

So, if $|x-a|<\delta_1$, the fact that $|f(x)-b|<\delta_2|$ implies $|g(f(x)-c|<\epsilon$. Therefore, $$\lim_{x\to a}g(f(x))=c.$$

Tim Raczkowski
  • 107
  • That's right but in many problems it can't be applied . – S.H.W Feb 01 '18 at 15:46
  • It should apply if the first two limits exist. – Tim Raczkowski Feb 01 '18 at 15:53
  • I mean in many substitutions we can't use that theorem . Undoubtedly , if the conditions of the theorem is fulfilled we can use that . – S.H.W Feb 01 '18 at 15:57
  • Yes, I guess I'm missing the point of your comment. Are you looking for some other answer? – Tim Raczkowski Feb 01 '18 at 15:58
  • In fact yes but anyway thanks for the answer . – S.H.W Feb 01 '18 at 15:59
  • OK, I'm not sure there is one. This is the principle that allows substitution in limits. Perhaps someone more knowledgeable may have deeper insights. :) – Tim Raczkowski Feb 01 '18 at 16:01
  • It's very challenging to me that use substitution without any proof . I appreciate your help , thanks a lot . :) – S.H.W Feb 01 '18 at 16:05
  • Have you seen the proof of the theorem I mentioned. If not , I can provide one. – Tim Raczkowski Feb 01 '18 at 16:11
  • Yes I do , it's really hard to grasp it ! – S.H.W Feb 01 '18 at 16:12
  • OK, I'll edit my answer – Tim Raczkowski Feb 01 '18 at 16:14
  • 1
    The above answer is correct, and can be complemented with those provided in the posts indicated by H. Lundmark. To help you visualize that, consider to graph $g(x)$ on a rubber sheet. Then imagine the sheet subject to a deformation along the $x$ axis, given by $f(x)$. If the deformation does not cause a tear in the sheet ... can you see what's going on ? – G Cab Feb 11 '18 at 17:40
  • 1
    This answer is not quite right. For example, suppose $f(x) = x\sin (1/x)$ and $g(x) = 1$ for all $x\ne 0,$ $g(0)=0.$ We have $\lim_{x\to 0} f(x) = 0,$ and $\lim_{x\to 0} g(x)=1.$ But $\lim_{x\to 0}g\circ (x)$ fails to exist, oscillating between $0$ and $1$ in every neighborhood of $0.$ – zhw. Feb 13 '18 at 23:01
  • @zhw. is correct. $g$ must be continuous at $b$ in order for this to work. – domdrag Jan 27 '22 at 11:40
2

Let's discuss the case you mention: Suppose $f:(a,\infty)\to \mathbb R$ for some $a>0.$ Here is a basic substitution result for this case:

$$\tag 1\lim_{x\to \infty}f(x) = L\, \iff\, \lim_{x\to 0^+}f(1/x) = L.$$

Your question was: "is that true if we say : $\lim\limits_{x \to \infty } x - \lfloor x \rfloor $ doesn't exist so we conclude $\lim\limits_{x \to 0^+ } 1/x - \lfloor 1/x \rfloor $ also doesn't exist?" The answer is yes, by the result $(1).$ Let's go through the logic: Define $f(x) = 1/x - \lfloor 1/x \rfloor.$ Suppose $\lim\limits_{x\to \infty}f(x)$ fails to exist. Could $\lim\limits_{x\to 0^+}f(1/x)$ exist? No! If it did, then then $(1)$ would imply $\lim\limits_{x\to \infty}f(x)$ exists, contradiction. That's what the $\iff$ is all about.

So in fact $(1)$ tells us that $\lim\limits_{x\to \infty}f(x)$ fails to exist $\iff \lim\limits_{x\to 0^+}f(1/x)$ fails to exist.

Added later: General result on limits and substitution The following result covers many, but not all, cases of interest. (Missing is the case of one-sided limits.)

Thm: Let $I,J$ be open intervals. Let $s: I \to J$ be a a continuous bijection. Let $a\in I$ and set $b=s(a).$ Suppose $f: J\setminus\{b\} \to \mathbb R.$ Then $\lim_{y\to b} f(y) = L$ iff $\lim_{x\to a} f(s(x))=L.$

Notes: 1. $y=s(x)$ of course is our substitution. 2. The hypotheses imply $s^{-1}: J \to I$ is a continuous bijection. We need this for both directions, i.e., the "iff". 3. We have taken care to not assume $f$ is defined at $b.$ Hence also $f(s(x))$ is not defined at $a.$ This is important for limits in general, for example in the famous case of $\lim_{x\to 0} (\sin x)/x.$

Proof of Thm: $\implies:$ Let $\epsilon>0.$ Then there exists $\delta > 0$ such that $(b-\delta, b+ \delta)\subset J$ and $y\in (b-\delta, b+ \delta)\setminus \{b\}$ implies $f(y) \in (L-\epsilon,L+\epsilon).$

Since $s$ is continuous, there exists $\gamma > 0$ such that $(a-\gamma, a+ \gamma)\subset I$ and $x\in (a-\gamma, a+ \gamma)\setminus \{a\}$ implies $s(x) \in (b-\delta, b+ \delta)\setminus \{b\}.$ For such $x$ we have $f(s(x)) \in (L-\epsilon,L+\epsilon).$ Thus $\lim_{x\to a} f(s(x))=L$ as desired.

$\impliedby$ Briefly: This time we assume $f\circ s,$ which is defined on $I\setminus \{a\},$ has limit $L$ at $a.$ From our work above, this implies $(f\circ s)\circ s^{-1} \to L$ at $b.$ Since $(f\circ s)\circ s^{-1}=f,$ we have $f\to L$ at $b$ as desired.

zhw.
  • 105,693
  • Yes , please write the proof but this is an example of substitution . How we can prove it in the general case ? My mean is changing variable in all situation works ? (i.e $\lim_{x \to a} f(g(x)) \iff \lim_{x \to b} f(x) $ if $\lim_{x \to a} g(x) = b$ ) – S.H.W Feb 16 '18 at 11:00
  • @S.H.W The "all situations" result is not correct in the way you stated it. We need to be more careful. Did you read my comment to the other answer? – zhw. Feb 16 '18 at 20:20
  • Oh , I see . That's a nice example . Can you mention the conditions where we can use that without any problem ? – S.H.W Feb 16 '18 at 22:01
  • @S.H.W I've added a proof for a general result. – zhw. Feb 17 '18 at 17:26