Relation between lim x->infinity of f(x) and lim n->infinity of f(n)

[Ozma]

Assume [imath]\lim_{x \to \infty} f(x)=L \in \mathbb{R} \cup {-\infty,\infty}[/imath]. I know that then for any sequence [imath]x_n \to \infty[/imath] it is [imath]f(x_n)=L[/imath] as [imath]n \to \infty[/imath], hence in particular for the sequence [imath]x_n=n \to \infty[/imath] as [imath]n \to \infty[/imath] it is [imath]f(n) \to \infty[/imath] as [imath]n \to \infty[/imath]. So it should be true that [imath]\lim_{x \to \infty} f(x)=L \implies \lim_{n \to \infty} f(n)=L[/imath]. Is this correct?

What about the vice versa? I think it is false, because for example the indicator function
[math]\chi_{\mathbb{R} \setminus \mathbb{N}}=\begin{cases} 1, \ \text{if} \ x \in \mathbb{R} \setminus \mathbb{N} \\ 0, \ \text{if} \ x \in \mathbb{N} \end{cases}[/math]is, for [imath]n \in \mathbb{N}[/imath], such that [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(n)=0[/imath] and so its limit as [imath]n \to \infty[/imath] is [imath]0[/imath] while for [imath]a_n=n+1/2 \in \mathbb{Q}_{>0} \setminus \mathbb{N}[/imath] it is [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath] and so [imath]\lim_{n \to \infty} \chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath], making the limit as [imath]x \to \infty[/imath] of [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}[/imath] not existent.

Is this whole reasoning correc?

 

[blamocur]

Looks right to me. For the counterexample of the opposite statement you can also use [imath]f(x) = \sin (\pi x)[/imath].

 

[Zermelo]

Assume [imath]\lim_{x \to \infty} f(x)=L \in \mathbb{R} \cup {-\infty,\infty}[/imath]. I know that then for any sequence [imath]x_n \to \infty[/imath] it is [imath]f(x_n)=L[/imath] as [imath]n \to \infty[/imath], hence in particular for the sequence [imath]x_n=n \to \infty[/imath] as [imath]n \to \infty[/imath] it is [imath]f(n) \to \infty[/imath] as [imath]n \to \infty[/imath]. So it should be true that [imath]\lim_{x \to \infty} f(x)=L \implies \lim_{n \to \infty} f(n)=L[/imath]. Is this correct?

What about the vice versa? I think it is false, because for example the indicator function
[math]\chi_{\mathbb{R} \setminus \mathbb{N}}=\begin{cases} 1, \ \text{if} \ x \in \mathbb{R} \setminus \mathbb{N} \\ 0, \ \text{if} \ x \in \mathbb{N} \end{cases}[/math]is, for [imath]n \in \mathbb{N}[/imath], such that [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(n)=0[/imath] and so its limit as [imath]n \to \infty[/imath] is [imath]0[/imath] while for [imath]a_n=n+1/2 \in \mathbb{Q}_{>0} \setminus \mathbb{N}[/imath] it is [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath] and so [imath]\lim_{n \to \infty} \chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath], making the limit as [imath]x \to \infty[/imath] of [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}[/imath] not existent.

Is this whole reasoning correc?

But what do you think the extra conditions for the opposite statement are? I’m just curious ?

 

[lex]

Assume [imath]\lim_{x \to \infty} f(x)=L \in \mathbb{R} \cup {-\infty,\infty}[/imath]. I know that then for any sequence [imath]x_n \to \infty[/imath] it is [imath]f(x_n)=L[/imath] as [imath]n \to \infty[/imath], hence in particular for the sequence [imath]x_n=n \to \infty[/imath] as [imath]n \to \infty[/imath] it is [imath]f(n) \to \infty[/imath] as [imath]n \to \infty[/imath]. So it should be true that [imath]\lim_{x \to \infty} f(x)=L \implies \lim_{n \to \infty} f(n)=L[/imath]. Is this correct?

What about the vice versa? I think it is false, because for example the indicator function
[math]\chi_{\mathbb{R} \setminus \mathbb{N}}=\begin{cases} 1, \ \text{if} \ x \in \mathbb{R} \setminus \mathbb{N} \\ 0, \ \text{if} \ x \in \mathbb{N} \end{cases}[/math]is, for [imath]n \in \mathbb{N}[/imath], such that [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(n)=0[/imath] and so its limit as [imath]n \to \infty[/imath] is [imath]0[/imath] while for [imath]a_n=n+1/2 \in \mathbb{Q}_{>0} \setminus \mathbb{N}[/imath] it is [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath] and so [imath]\lim_{n \to \infty} \chi_{\mathbb{R} \setminus \mathbb{N}}(a_n)=1[/imath], making the limit as [imath]x \to \infty[/imath] of [imath]\chi_{\mathbb{R} \setminus \mathbb{N}}[/imath] not existent.

Is this whole reasoning correc?

Assume [imath]\lim_{x \to \infty} f(x)=L \in \mathbb{R} \cup \textcolor{red}{\{}-\infty,\infty\textcolor{red}{\}}[/imath]. I know that then for any sequence [imath]x_n \to \infty[/imath] it is [imath]f(x_n)=L[/imath] as [imath]n \to \infty[/imath], hence in particular for the sequence [imath]x_n=n \to \infty[/imath] as [imath]n \to \infty[/imath] it is [imath]f(n) \to \textcolor{red}{L}[/imath] as [imath]n \to \infty[/imath]. So it should be true that [imath]\lim_{x \to \infty} f(x)=L \implies \lim_{n \to \infty} f(n)=L[/imath]. Is this correct?