How to prove there is no limit of cos1/x using theorm of limit

[flying2000]

How to prove there is no limit(x->0) of cos1/x using theorem of limit?
Anybody can give me some hints?
thanks

 

[Dr-NiKoN]

I'm just learning this myself, but here is how I would approach it.

$\lim_{x\to 0}f(x)=\cos(\frac{1}{x})$
You know what happens to $\lim_{x\to 0}f(x) = \frac{1}{x}$

You also know what a graf with $f(x) = \cos(x)$ looks like.

Now consider

$y = \frac{1}{x}$

and

$\lim_{y\to\infty}f(y)=\cos(y)$

 

[flying2000]

thanx

Choose an arbitrary number called "L". (Think of it as between -1 and 1 if you like)
Show that there exists a neighbourhood U of 0, so that given |L|<1, there will always be some x in U so that |L-cos(1/x)|>1/2
but I don't know how to continue...

I'm just learning this myself, but here is how I would approach it.

$\lim_{x\to 0}f(x)=\cos(\frac{1}{x})$
You know what happens to $\lim_{x\to 0}f(x) = \frac{1}{x}$

You also know what a graf with $f(x) = \cos(x)$ looks like.

Now consider

$y = \frac{1}{x}$

and

$\lim_{y\to\infty}f(y)=\cos(y)$

 

[flying2000]

my question is how to find x to makes |L-cos(1/x)|>1/2

my question is how to find x to makes |L-cos(1/x)|>1/2

 

[matt grime]

Use the easier, and equivalent, definition of limit.

If you can find two sequences a(n) and b(n) tending to zero such that cos(a(n)) and cos(b(n)) tend to different numbers you're done.

 

[flying2000]

thanx,but I want to know if I can prove it by E-Delta therom? Hope U can Help me..

thanx,but I want to know if I can prove it by E-Delta therom? Hope U can Help me..

Use the easier, and equivalent, definition of limit.

If you can find two sequences a(n) and b(n) tending to zero such that cos(a(n)) and cos(b(n)) tend to different numbers you're done.

 

[jebeagles]

Take limit as x approaches 0 from the right (lim->0+) of cos(1/x). Using direct substitution, you would get cos(infinity). cos(infinity) does not approach a single value, because it is not a monotonic function over the required interval. So, no limit exists, because no single value is approached. If one side of a limit does not exist, the limit does not exist.

This is the way I learned to do limits. Not sure if it applies here though.

 

[matt grime]

To prove the counter example it suffices to show that given any d there is a e such that there is some x with |cos(x)-L| > e and |x|<d

but this is trivial. firstly L must be between -1 and 1, and let x be some sufficiently large solution to cos(x)=1 or -1, then one of |1-L| and |-1-L| must be greater than 1 (which we can choose to be e)

nb e:=epsilon, d:=delta

 

[flying2000]

thanx

I got it now,thanx a lot
Actually,I thought it should hold for all x ,my mistake!

To prove the counter example it suffices to show that given any d there is a e such that there is some x with |cos(x)-L| > e and |x|<d

but this is trivial. firstly L must be between -1 and 1, and let x be some sufficiently large solution to cos(x)=1 or -1, then one of |1-L| and |-1-L| must be greater than 1 (which we can choose to be e)

nb e:=epsilon, d:=delta

 

[matt grime]

In the negation of propositions never forget that 'for all' is changed to 'there exists'