Condition for $\displaystyle \lim_{x\to 0}f(x)=\lim_{n\to\infty}f\left(\frac{1}{n}\right)$?

Question

Let $f:{\Bbb R}\to{\Bbb R}$. Is there a courterexample for the following equality or is it always true?

$$\lim_{x\to 0}f(x)=\lim_{n\to\infty}f\left(\frac{1}{n}\right)$$

What I think is that one might need a non-continuous function since this is always true for a continuous function. Would $1_{\Bbb Q}$ work? Are there any other counterexamples?

$1_\mathbb{Q}$ works in the sense that the RHS limit exists and the LHS limit does not. There are plenty of other counterexamples in this sense (e.g. the one in my answer.) — Trevor Wilson, Feb 16 '13 at 00:11
@SalechAlhasov: It is the indicator function for the rational numbers, i.e. $f(x)=1$ if $x\in\Bbb Q$ and $0$ otherwise. — Clayton, Feb 16 '13 at 04:32

Clayton · Accepted Answer · 2013-02-16T04:33:10.803

7

Actually, even for a function which is discontinuous at only one point, this need not be true. For example: $$f(x)=\begin{cases}\sin(\pi/x)&:x\neq0\\ 0&:x=0\end{cases}.$$ This equals $0$ for every $n\in\Bbb Z\setminus\{0\}$ as $\pi/(1/n)=n\pi$ but the limit as $x\to0$ does not exist. However, if the function is continuous, then yes, the limits will be equal.

Thus, as you can see, the only way to guarantee the limits are equal is if $$\lim_{x\to0}f(x)$$ exists.

edited Feb 16 '13 at 04:33

answered Feb 16 '13 at 00:07

Clayton

24,751

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What does "subsequence of $x \to 0$" mean? – Trevor Wilson Feb 16 '13 at 00:31
@TrevorWilson: When considering $x\to0$, it's equivalent to considering all possible sequences converging to $0$; this is just one of those sequences. – Clayton Feb 16 '13 at 00:52
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I know that sequential continuity is equivalent to continuity in $\mathbb{R}$, I'm just questioning your notation. To me the phrase "$x \to 0$" does not denote a sequence, so I don't know what a subsequence of it would be. – Trevor Wilson Feb 16 '13 at 00:55
Nice example! +1 – user1551 Feb 16 '13 at 04:26
Nice function, the beautiful one on Earth +1 – Mikasa Feb 17 '13 at 19:43

score 5 · Answer 2 · answered Feb 16 '13 at 00:07

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If the left hand side limit exists, then both limits exist and are equal. However, it is possible for the right hand side limit to exist and the left hand side limit not to exist (in which case $f$ must be discontinuous.) For example, let $f(x) = 1$ if $x = 1/n$ for some $n \in \mathbb{N}$ and $f(x) = 0$ otherwise.

answered Feb 16 '13 at 00:07

Trevor Wilson

16,989

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I'm assuming the limit on the right hand side is taken over $n \in \mathbb{N}$, by the way. – Trevor Wilson Feb 16 '13 at 00:09
@blf Then the limit on the RHS would be more commonly written as $\lim_{x\to 0^+} f(x)$, and the answer is the same (but you would need a different counterexample, like the indicator function for $[0,\infty)$.) – Trevor Wilson Feb 16 '13 at 00:14

Condition for $\displaystyle \lim_{x\to 0}f(x)=\lim_{n\to\infty}f\left(\frac{1}{n}\right)$?

2 Answers2