How to show lim sin(1/x) DNE as x->0

[eckiller]

How to show lim sin(1/x) DNE as x-->0

Hello,

I am trying to see how lim sin(1/x) does not exist as x-->0. It is obvious from a graph.

I think one way to do this is to pick two sequences converging to 0 and show that the limit of these sequences do not equal each other. For example, I can pick a sequence where sin gives me +1 and another one where sin gives me -1. I think this would work. Can someone confirm that this strategy is sound?

Also, I would also like to see how to show this using epsilon-delta argument at well.

Thanks in advance.

 

[HallsofIvy]

Yes, that will work. If the limit of a continuous function exists at a: that is, if
lim_{x->a} f(x) exists, then for any sequence {x_n}, lim_{n->\inf }f(x_n) exists and is the same. If, for any such sequence the limit does not exist, or if two such sequences do not have the same limit, then the limit itself does not exist.

 

[M98Ranger]

To show that lim sin(1/x) does not exist as x-->0, we can use the sequential definition of a limit. As you mentioned, we can choose two sequences, {x_n} and {y_n}, both converging to 0, but with different values for sin(1/x_n) and sin(1/y_n). For example, let x_n = 1/n and y_n = 1/(2n+1). As n approaches infinity, x_n and y_n both converge to 0, but sin(1/x_n) alternates between 1 and -1, while sin(1/y_n) alternates between 1 and 0. This shows that the limit of sin(1/x) as x-->0 does not exist.

To use the epsilon-delta argument, we can assume that the limit of sin(1/x) exists as x-->0 and try to find a contradiction. Let's say the limit is L, then for any ε>0, there exists δ>0 such that |sin(1/x) - L|<ε whenever 0<|x|<δ. However, we can choose x = 1/(2n+1) and y = 1/n, both within δ, but with different values for sin(1/x) and sin(1/y). This contradicts the assumption that the limit exists, thus showing that lim sin(1/x) does not exist as x-->0.

I hope this helps! Let me know if you have any further questions.