# Limits and infinite oscillations

Hello everyone,

I am having trouble understanding the concept of a limit not existing for functions like sin (1/x) when x tends to 0. The good book says that the function "does not settle on any value as we get closer to x" implying some infinite oscillation. I am having trouble visualizing it and why it should happen with this particular function.

Any kind soul here willing to elaborate on this and help me understand this better? I would be extremely grateful.

Many thanks,

Luca

## Answers and Replies

dx
Homework Helper
Gold Member
Think about sin(x) as x goes to infinity. Do you know what the graph of sin(x) looks like? Does it settle down to any particular value as x goes to infinity?

Hi there,

Thanks for the reply. The sine graph is oscillating between -1 and 1 every with a period of 2 pi. How would this change into infinite oscillations for sin(x) as x approaches infinity...

Thanks,

Luca

Ok, I think I understand this better now as we approach infinity, the value will keep oscillating between -1 and 1 and there is no one value where the function will settle down to...fair enough :)

Thanks,

Luca

dx
Homework Helper
Gold Member
You said it yourself. It oscillates between 0 and 1. Just draw the graph of sin. As you go further and further in the positive direction of the x axis, does it settle down to some value? No, it keeps oscillating between 1 and 0.

EDIT: Sorry, I don't know what I was thinking. I meant to say oscillating between 1 and -1.

Last edited:
HallsofIvy
$sin(n\pi)= 0$ for all n. Let $x_n= 1/(n\pi)$. Then as n goes to infinity, $x_n$ goes to 0 and $sin(1/x_n)= 0$ for all n.
But $sin(2n\pi+ \pi/2)= 1$ for all n. Let $x_n= 1/(2n\pi+ \pi/2)$. Then as n goes to infinity, $x_n$ goes to $sin(1/x_n)= 1$ for all n.
If $\lim_{x\rightarrow 0} sin(1/x)$ existed, the limit of $sin(x_n)$ would have to be that value for any sequence $x_n$ converging to 0. Since the two sequences above have different limits, the limit of the function cannot exist.