Limits and infinite oscillations

[pamparana]

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

 

[dx]

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?

 

[pamparana]

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

 

[pamparana]

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]

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.

 

[HallsofIvy]

Another way to look at it:

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.