Function oscillates from -1 to 1 so limit

[zorro]

Why is it that

does not exist?
I read in books that the function oscillates from -1 to 1 so limit does not exist. I don't understand how can we conclude that from the given reason?
On the other hand,

exists
Please explain me the difference between the two.

 

[dacruick]

hi there,

you're right that sine oscillates. if x goes to 0, that means that your term inside will be infinity. This means that the sin(1/x) could be anywhere between -1 and 1, and is therefore undefined.

but if you put the x in front, sin(1/x) can be anywhere between -1 and 1, and the limit will still be 0 right?

 

[zorro]

but if you put the x in front, sin(1/x) can be anywhere between -1 and 1, and the limit will still be 0 right?

sin(1/x) can be anywhere between -1 and 1
so xsin(1/x) can be anywhere between -x and x
Is'nt it?

 

[dacruick]

right. And if x goes to 0 then...

 

[zorro]

ok I got it
Thanx :)

 

[Mark44]

hi there,

you're right that sine oscillates. if x goes to 0, that means that your term inside will be infinity.

Technically, this isn't correct.
\lim_{x \to 0}}\frac{1}{x}
doesn't exist at all, in any sense. The left- and right-side limits are different.

 

[zorro]

Technically, this isn't correct.
\lim_{x \to {\infty}}\frac{1}{x}
doesn't exist at all, in any sense. The left- and right-side limits are different.

Did you mean sin(1/x) ?
If yes then how can you prove that left and right -side limits are different?

 

[Mark44]

Did you mean sin(1/x) ?
If yes then how can you prove that left and right -side limits are different?

Sorry, I wrote something different from what I was thinking.

This is what I meant:
\lim_{x \to 0} \frac{1}{x}

That limit doesn't exist, in any sense.

 

[zorro]

Sorry, I wrote something different from what I was thinking.

This is what I meant:
\lim_{x \to 0} \frac{1}{x}

That limit doesn't exist, in any sense.

But \lim_{x \to 0} \frac{1}{x} is infinity. It does exist.
lol were you again thinking something else while writing?

 

[Mark44]

But \lim_{x \to 0} \frac{1}{x} is infinity. It does exist.

This limit is NOT infinity, and absolutely does NOT exist! The left- and right-side limits are as different as it is possible to be, which means that the two-sided limit doesn't exist.

 

[dacruick]

Mark is right, that limit doesn't exist.

But I don't really understand the relevancy of this

 

[Mark44]

you're right that sine oscillates. if x goes to 0, that means that your term inside will be infinity. This means that the sin(1/x) could be anywhere between -1 and 1, and is therefore undefined.

Mark is right, that limit doesn't exist.

But I don't really understand the relevancy of this

I was responding to something you said (first quote above). You said that the "term inside" (which I understood to mean 1/x) will be infinity as x approaches 0. That's not true, since the limit of 1/x as x approaches 0 doesn't exist.

 

[dacruick]

I was responding to something you said (first quote above). You said that the "term inside" (which I understood to mean 1/x) will be infinity as x approaches 0. That's not true, since the limit of 1/x as x approaches 0 doesn't exist.

Thats true, my mistake. Mathematically that's incorrect but thinking about it that way has always helped me feel through these questions.

 

[Mark44]

It would be more helpful to distinguish between limits that are infinite and limits that don't exist at all. For example, \lim_{x \to \infty} e^x = \infty
but \lim_{x \to 0} \frac{1}{x} doesn't exist at all.

 

[dacruick]

It would be more helpful to distinguish between limits that are infinite and limits that don't exist at all. For example, \lim_{x \to \infty} e^x = \infty
but \lim_{x \to 0} \frac{1}{x} doesn't exist at all.

I think what's really happening is in that case I just say that x approaches zero from the right or left and then the limit exists