How to find lim as x -> inf. of sin(x) algebraically

[PhizKid]

Homework Statement

\lim_{x \to \infty } \sin x

Homework Equations

The Attempt at a Solution

I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.

 

[Bacle2]

Homework Statement

\lim_{x \to \infty } \sin x

Homework Equations

The Attempt at a Solution

I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.

The usual definition of limit as x→∞ is:

\lim_{x \to \infty } \f(x) =L

if, given ε>0 , there is an M such that for all x>M , |f(x)-L|<ε

So , algebraically, you can show the limit does not exist by choosing the right value

of ε for which the def. is not satisfied.

 

[HallsofIvy]

You can't "find the limit of sin(x) as x goes to infinity" because it does not have a limit. No matter how large x is, the will be larger values for which sin(x) has values of -1, 1, etc. sin(cx) does NOT get close to anyone number as x goes to infinity.

 

[Bacle2]

But PhizKind is not trying to find the limit,but instead to show , algebraically,

that the limit does not exist.

 

[Ray Vickson]

Homework Statement

\lim_{x \to \infty } \sin x

Homework Equations

The Attempt at a Solution

I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.

The function sin(x) is periodic, with period 2π. That means that sin(x+2kπ) = sin(x) for k = ±1, ±2, ±3, ... for any x. So, the graph of y = sin(x) for x between 10,000,000π and 10,000,002π is an exact copy of the graph of y = sin(x) for x between 0 and 2π.

RGV

 

[STEMucator]

You need to use a formal definition for this question. In particular :

\forallM>0, \existsN | x>N \Rightarrow f(x)>M

Then start with f(x) > M and massage it until you can find your suitable N.

 

[Bacle2]

You need to use a formal definition for this question. In particular :

\forallM>0, \existsN | x>N \Rightarrow f(x)>M

Then start with f(x) > M and massage it until you can find your suitable N.

I think you are referring here to the case where the limit is ∞, not the limit as x→∞.

For f(x)>1 , you willfind no M with the property you described.

 

[Bacle2]

Homework Statement

\lim_{x \to \infty } \sin x

Homework Equations

The Attempt at a Solution

I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.

Maybe you can do this by contradiction:

If the limit existed, and it was equal to L, you would be able to say that, when you

go far-enough (i.e., when x is largerthan M ), you can get to within any level of accuracy

(i.e., within ε ) to the limit.

This means, formally , that there is some M>0 with:

|Sinx-L|<ε when x>M

But this will not happen using , e.g., Ray Vickson's argument. You want:

-ε< Sinx-L <ε

But no matter what value of L you choose, L will oscillate.

 

[STEMucator]

I think you are referring here to the case where the limit is ∞, not the limit as x→∞.

For f(x)>1 , you willfind no M with the property you described.

Ah yes I forgot about that, I meant to say |f(x)-L| < M, but using ε in this case would have been more appropriate than using M ( Even though it really doesn't matter just some formalism ).