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Using the definition of limits, how can I prove that $f(x)=\sin(x)$ has no limit as $x \rightarrow\infty$?

rschwieb
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Guest11235
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    Try evaluating $\sin x$ at $ x = \frac{\pi}{2} + k \pi$ for each $k\in\mathbb{N}$, that should get your started. – James Oct 28 '14 at 14:16
  • or $x_n=2k\pi$ with $n \in \mathbb{N}$ and $k \in \mathbb{N}$ – Dr. Sonnhard Graubner Oct 28 '14 at 14:18
  • @Dr.SonnhardGraubner $\sin 2k\pi = 0$ for each $k\in\mathbb{N}$, this, by itself, isn't going to help. Also, your sequence $x_n$ is constant. – James Oct 28 '14 at 14:19
  • yes i know it and $x_n$ goes to infinity for $k$ goes to infinity, we want to prove that es limit doesn't exist, you must take another sequence e.g. $x_n=\frac{\pi}{2}+k\pi$ like above – Dr. Sonnhard Graubner Oct 28 '14 at 14:23

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It cannot have a limit that is $\ge 0$, because there are arbitrarily large $x$ with $\sin x=-1$.

It cannot have a limit that is $\le 0$, because there are arbitrarily large $x$ with $\sin x=1$.

hmakholm left over Monica
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take two sequences $x_n=2n\pi$ thus you will get $\lim_{n \to \infty}\sin(2n\pi)=0$ or $\lim_{n\to \infty}\sin(\pi/2+2n\pi)=1$

Dr. Sonnhard Graubner
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Perhaps your issue in "using the definition" is dealing with the $\epsilon$-$\delta$ formulation of the statement.

Hint: We can write "the limit at infinity does not exist" as

There is no $L \in \Bbb R$ that qualifies as a limit. That is, for every $L$: there exist an $\epsilon > 0$ such that for any $x_0 > 0$, we never have $$ |\sin(x) - L| < \epsilon \text{ when } x > x_0 $$ which is to say that for every $x_0$, there is an $x > x_0$ such that $$ |\sin(x) - L| \geq \epsilon $$

Ben Grossmann
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