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My motivation was actually coming from :Change of Variables in Limits (Part 1)

and the counter example:

For $g(x)=x\sin(1/x)$, and $f(x)=\left\{\begin{aligned}&2,\quad x= 0\\&1,\quad \text{otherwise}\end{aligned}\right.$.

We have $\displaystyle \lim_{x\to 0}f(g(x))$ doesn't exist but $\displaystyle\lim_{y\to 0} f(y)=1$ so they are not equal.

but in the case of $\lim\limits_{x\to a}\dfrac{\sin(g(x))}{g(x)}$ where $\lim\limits_{x\to a}g(x)=0$ we always say this former limit is $1$.

If we can think the continuous extensions $\tilde f$ of the functions $f$ which has removable discontinuity at that point. I can show $$\lim_{x\to a}\tilde f(g(x))=\lim_{x\to 0}f(x)$$ where $\lim\limits_{x\to a}g(x)=0$.By taking limit inside using continuity of $\tilde f$. Of course $\lim\limits_{x\to 0} f(x)$ is assumed to exist (that's why it has only removable discontinuity).

In the case of $\;\tilde {\dfrac {\sin(x)}{x}}=\begin{cases}\dfrac {\sin(x)}{x}&\text{if}& x\neq 0\\ 1 &\text{if} &x=0\end{cases}$

However as you can see also from the very above link $\lim\limits_{x\to a}\tilde f(g(x))\neq\lim\limits_{x\to a}f(g(x))$ in general. So there might be a function $g$ that satisfies:

$\lim\limits_{x\to a}\dfrac{\sin(g(x))}{g(x)}\neq1=\lim\limits_{x\to 0}\dfrac{\sin(x)}{x}$ where $\lim\limits_{x\to a}g(x)=0$

Angelo
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Micheal Brain Hurts
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  • No it can happen as you can see from the link i presented. As $g=xsin(1/x)$ is defined around $(-a,a)\setminus{0}$ – Micheal Brain Hurts Oct 19 '24 at 23:33
  • The same problem occurs with your quotient issue. What if $g(x)=0$ infinitely often as $x\to a$? Then the limit is not defined by most definitions of limit — you have to exclude the points where $g(x)=0$. However, some textbooks/people modify the definition of the limit appropriately. – Ted Shifrin Oct 19 '24 at 23:33
  • I don't think you read it careful enough. Since there are not iff statement for these change of variable problems I wanted to understand what is the difference between $sin(x)/x$ and $f(x)=\left{\begin{aligned}&2,\quad x= 0\&1,\quad \text{otherwise}\end{aligned}\right.$ using the same change of variable $g=xsin(1/x)$ as $g$ passes through 0 infinitely many times as you said @TedShifrin – Micheal Brain Hurts Oct 19 '24 at 23:41
  • I don't think you read what I wrote carefully enough. I'm saying that — according to a standard calculus textbook definition — $\lim\limits_{x\to 0} \dfrac{\sin(g(x))}{g(x)}$ does not exist, despite the fact that $g(x)\to 0$ as $x\to 0$. – Ted Shifrin Oct 19 '24 at 23:46
  • Ok then we talk specifically about $g(x)=xsin(1/x)$ but why $\lim\limits_{x\to 0} \dfrac{\sin(g(x))}{g(x)}$ doesnot exists? Since standards changing as you say, which point in the standard definition makes it doesnot exist? Because I used desmos to graph $\dfrac{\sin(xsin(1/x))}{xsin(1/x)}$ in which around 0 its limit is 1. @TedShifrin – Micheal Brain Hurts Oct 19 '24 at 23:49
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    We're talking about mathematics, not about Desmos :) Desmos ignores the places where the denominator is $0$. You need to edit your post to include a careful definition of LIMIT as you're using it. Most definitions will require that the domain of the function includes the set $0<|x-a|<\delta$ for some $\delta>0$. – Ted Shifrin Oct 19 '24 at 23:58
  • yeah they should be used otherwise when there is a composition like this, as you already said we cannot arrange any neighborhood that $0<|...|<\delta$. Thank you – Micheal Brain Hurts Oct 20 '24 at 00:05

2 Answers2

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The function $g(x) = 0$ works since ${\displaystyle \lim_{x \rightarrow a} \frac{\sin(g(x))}{g(x)}}$ doesn't exist; it's not defined. Any other function where $g(x) = 0$ arbitrarily close to $x = a$ will have the same issue. Otherwise the limit will be $1$. There's no way the limit can exist and not be $1$.

You can change the domain of ${\displaystyle \frac{\sin(g(x))}{g(x)}}$ to exclude the points where $g(x) = 0$, and if you're left with points arbitrarily close to $x = a$, then you can view the limit as being on the domain $\{x: g(x) \neq 0\}$, in which case the limit is $1$ once again.

Alternatively you can replace ${\displaystyle \frac{\sin x}{x}}$ by its continuous extension to $x = 0$, and then once again the limit will exist and equal $1$, as you are taking the limit of the composition of two continuous functions.

Zarrax
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1

Let $h(x)$ be a function continuous at $0$ and $g(t)$ a function so that $\lim_{t\to a}g(t)=0.$ Then for any sequence $t_n\neq a,$ $t_n\to a,$ we have $g(t_n)\to 0.$ Hence $\lim_{n\to \infty}h(g(t_n))= h(0).$ By the Heine definition of the limit we obtain $\lim_{t\to a}h(g(t))=h(0)$ . Apply this to $$h(x)=\begin{cases} {\sin x\over x} &x\neq 0\\ 1& x=0\end{cases}$$ to get $\lim_{t\to a}h(g(t))=1.$

Ryszard Szwarc
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