Thread replies: 8
Thread images: 1
Thread images: 1
File: 1499517752239458952.gif (131KB, 290x419px) Image search:
[Google]
131KB, 290x419px
Let [math]f:\mathbf{R}\to\mathbf{R}[/math] be some function and [math]a_n[/math] a sequence such that for every [math]n[/math] [math]a_n=f(n)[/math]. How to show that [math]lim_{n\to\infty} a_n=lim_{x\to\infty}f(x)[/math], where on the LHS [math]n[/math] ranges over naturals and on the RHS [math]x[/math] ranges over reals?
>>
>>9095101
yes.
>>
>>9095101
You can't because that's wrong.
>>
>>9095101
Consider [math]f(x) = \sin(x \pi)[/math].
Then [math]a_n = 0[/math] for all [math]n[/math] but [math]lim_{x\to\infty}f(x)[/math] does not exist.
>>
>>9095274
Yes. [math]f_\infty = \mathrm{lim}_{x \to \infty} f (x) [/math] exists if and only if [math]\mathrm{lim}_{n \to \infty} f(x_n) [/math] exists and is equal to [math]f_\infty[/math] for every sequence [math]x_n[/math] with [math]x_n \to \infty [/math] for [math] n \to \infty[/math].
>>
>>9095294
Ok, but , assuming both limits exist, [math]\lim_{n\to\infty} a_n=L[/math] doesn't imply [math]lim_{x\to\infty}f(x)=L[/math] right? Because all we know is that [math]\mathrm{lim}_{n \to \infty} f(x_n)=L[/math] for one sequence, namely [math]x_n=n[/math]
>>
>>9095308
It does imply that. If we assume both limits exist then we already know that [math]\mathrm{lim}_{n \to \infty} f(x_n) [/math] exists and is equal to some [math] f_\infty [/math] for every [math] x_n [/math] with [math] x \to \infty [/math]. As such we only have to check one of these sequences (namely [math] x_n = n [/math]) to find it and see that it's equal to [math] \mathrm{lim}_{n \to \infty} a_n [/math].
Thread posts: 8
Thread images: 1
Thread images: 1