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As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined.

Which opinion is correct?

Dylan C. Beck
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Soling
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    You reduce your expression then you compute the limit – B E I R U T Feb 02 '21 at 04:00
  • I think the limit is equal to $1$. you can draw the graph of your function, which is a straight line ($y=1$) and there are infinite number of holes for $x=\ldots,-2\pi,-\pi,0,\pi,2\pi,\ldots$ where at these points the limit is still equal to $1$. – Etemon Feb 02 '21 at 04:00
  • The complication you're encountering reminds me of similar complications encountered in multivariable calculus when evaluating a limit like $$\lim_{(x,y)\rightarrow (0,0)} \frac{\sin(x+y)}{x+y}$$ Some textbooks say this limit DNE because $f(x,y)=\frac{\sin(x+y)}{x+y}$ isn't defined on any open neighborhood containing the origin; other textbooks say this limit is equal to $1$ as one should only consider paths tending to the origin that don't intersect $x+y=0$. – Matthew H. Feb 02 '21 at 04:01
  • When evaluating limits, you don't need to care about the points outside the domain. – 49328481 Feb 02 '21 at 04:05
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    Your function is a function that is defined for all $x\neq n\pi$, $n\in\mathbb{Z}$. The value of the function where defined is $1$, but it is not everywhere defined on any interval of the form $(R,\infty)$. If your definition of limit requires the function to be defined on one such interval, the limit does not exist. If your definition of limit allows you to take the limit at $\infty$ provided only that for all $M$ there exist $x\gt M$ where the function is defined, then the limit is $1$. – Arturo Magidin Feb 02 '21 at 04:05
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    @49328481: So, I can compute $\lim_{x\to -1}\sqrt{x}$? – Arturo Magidin Feb 02 '21 at 04:06
  • @ArturoMagidin, great question and I'm not sure, maybe I was wrong. But that's the definition of limit I was thinking: $\forall \varepsilon \geq 0 \exists \delta \geq 0 \forall x \in D(f); |x - x_0| \leq \delta \implies |f(x) - L| \leq \varepsilon$. But now you make me think and when your $x$ is not in the domain, the expression would be vacuously true for any $L$. Is that right? – 49328481 Feb 02 '21 at 04:12
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    @49328481: See my answer: the definition you give is a common one, but there is an extra assumption that you are missing: $x_0$ is required to be an accumulation point of $\mathrm{dom}(f)$. Otherwise, your definition would allow the limit to exist and have all values at points that are not accumulation points of the domain. And while we are okay with a limit not existing, generally we do not want limits of real-valued function to have multiple values. You don’t want the limit to exist and be both $1$ and $-7$. – Arturo Magidin Feb 02 '21 at 04:18
  • @ArturoMagidin understood! Thanks! – 49328481 Feb 02 '21 at 04:22

2 Answers2

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It depends on exactly what your definition of limit is.

In calculus, limits of real-valued functions are often defined in such a way that in order for the limit as $x\to a$ to exist, the function must be defined on some interval of the form $(a-\delta,a+\delta)$, except perhaps at $a$. For limits as $x\to\infty$, you require the function to be defined on some interval of the form $(M,\infty)$. The definition using $\epsilon$s would look something like:

$\lim_{x\to\infty} f(x)=L$ if and only if for all $\epsilon\gt 0$ there exists $M$ such that if $x\gt M$, then $|f(x)-L|\lt\epsilon$.

However, sometimes limits are defined differently (slightly more generally, applicable to more functions), and the prerequisite for the limit as $x\to a$ to exist is that $a$ be an “accumulation point” of the domain of $f$. That is, that for all $\delta\gt 0$, there exist points $b$ in the domain of $f$ such that $0\lt |b-a|\lt \delta$. For limits as $x\to\infty$, one requires that for all $M$ there exist $b$ in the domain of $f$ such that $b\gt M$.

Here, the definition using $\epsilon$s would be:

$\lim_{x\to\infty} f(x)=L$ if and only if (i) for every $M$ there exists $x\gt M$ such that $x\in\mathrm{dom}(f)$; and (ii) for every $\epsilon\gt 0$ there exists $N$ such that if $x\gt N$ and $x\in\mathrm{dom}(f)$, then $|f(x)-L|\lt\epsilon$.

If your definition of limits is of the first type, then the limit you write does not exist, because the function does not meet the requirement to be able to even talk about a potential limit as $x\to\infty$: it is not defined on any interval of the form $(M,\infty)$, so you can always find an $x$ with $x\gt M$, and where $|f(x)-1|\lt\epsilon$ fails because the expression doesn’t even make sense. On the other hand, if your definition of limit is of the second type, then the limit is $1$, since the satisfies both conditions of that definition.

In short, neither opinion is right and neither is wrong: it depends on exactly what the symbol $\lim_{x\to\infty}f(x)=L$ means.

Arturo Magidin
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    Beautifully put. +1 – Dylan C. Beck Feb 02 '21 at 04:18
  • Are you allowed to say that $\lim_{x \to 0}x\log x = 0$? $\log x$ is undefined for $x<0$, and so it seems that you can only say $\lim_{x \to 0+}x\log x = 0$. Or is this another case where different authors adopt different conventions about how they define limits? – Joe Feb 18 '21 at 15:09
  • @Joe: Under the usual calculus convention, the regular limit does not exist because the function is not defined in an open neighborhood of $0$. So, yes, it depends on exactly how you define limits. – Arturo Magidin Feb 18 '21 at 15:52
  • Ironically, neighbourhoods are another example of whether different authors adopt slightly different conventions :) Of course, there's no actual ambiguity here. Thanks for this answer, Arturo. – Joe Feb 18 '21 at 21:37
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By the familiar definition (i.e., the "Wikipedia" definition), for a real-valued function $f(x),$ we have that $\lim_{x \to \infty} f(x)$ is a real number $L$ (if it exists) such that for every real number $\varepsilon > 0,$ there exists a real number $M$ such that for all $x > M,$ we have that $|f(x) - L| < \varepsilon.$

Like you have mentioned, if $x$ is not an integer multiple of $\pi,$ we have that $f(x) = \frac{\sin x}{\sin x} = 1$; however, for any integer $n,$ we have that $f(n \pi)$ is not defined. What can you conclude?

Dylan C. Beck
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  • I thought $x = n\pi$ is to be "skipped" because $f(x)$ can't be definied if $x$ is not in a domain of a function(in this case $x = n \pi$). – Soling Feb 02 '21 at 04:20
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    In the "familiar" definition of the limit of $f(x)$ at infinity, we need $f(x)$ to be defined for all $x > M$ for some real number $M$ sufficiently large. In this case, you cannot "skip" the values $x = n \pi$ for large $n.$ It is true that $\lim_{x \to n \pi} \frac{\sin x}{\sin x} = 1$ for every integer $n,$ but that is also a consequence of the definition of $\lim_{x \to a} f(x).$ See Professor Magidin's answer above for the complete picture. – Dylan C. Beck Feb 02 '21 at 04:25