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

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In summary, to prove that there is no limit as x approaches 0 of cos(1/x) using the limit theorem, we can provide a counterexample by showing that for any given d, there exists an e such that there is an x with |cos(x)-L| > e and |x| < d. This can be easily done by considering large solutions to cos(x)=1 or -1, which will make either |1-L| or |-1-L| greater than 1, satisfying the condition.
  • #1
flying2000
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How to prove there is no limit(x->0) of cos1/x using theorm of limit?
Anybody can give me some hints?
thanks
 
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  • #2
I'm just learning this myself, but here is how I would approach it.

[itex]\lim_{x\to 0}f(x)=\cos(\frac{1}{x})[/itex]
You know what happens to [itex]\lim_{x\to 0}f(x) = \frac{1}{x}[/itex]

You also know what a graf with [itex]f(x) = \cos(x)[/itex] looks like.

Now consider

[itex]y = \frac{1}{x}[/itex]

and

[itex]\lim_{y\to\infty}f(y)=\cos(y)[/itex]
 
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  • #3
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...

Dr-NiKoN said:
I'm just learning this myself, but here is how I would approach it.

[itex]\lim_{x\to 0}f(x)=\cos(\frac{1}{x})[/itex]
You know what happens to [itex]\lim_{x\to 0}f(x) = \frac{1}{x}[/itex]

You also know what a graf with [itex]f(x) = \cos(x)[/itex] looks like.

Now consider

[itex]y = \frac{1}{x}[/itex]

and

[itex]\lim_{y\to\infty}f(y)=\cos(y)[/itex]
 
  • #4
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
 
  • #5
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.
 
  • #6
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..
matt grime said:
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.
 
  • #7
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.
 
  • #8
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
 
  • #9
thanx

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

matt grime said:
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
 
  • #10
In the negation of propositions never forget that 'for all' is changed to 'there exists'
 

1. What is the theorem of limit?

The theorem of limit, also known as the limit theorem, is a fundamental concept in calculus that states if a function f(x) approaches a specific value L as x approaches a certain value c, then we can say that the limit of f(x) as x approaches c is equal to L. This is denoted as "lim f(x) = L" or "f(x) → L" as x → c.

2. How does the theorem of limit apply to cos1/x?

The theorem of limit can be used to determine the limit of any function, including the cosine function, as long as the function satisfies the conditions for the theorem to be applied. In the case of cos1/x, the limit can be evaluated as x approaches 0, since this is the only point where the function does not exist.

3. Can the limit of cos1/x be proven using the theorem of limit?

Yes, the limit of cos1/x can be proven using the theorem of limit. By applying the definition of the limit, we can show that the limit of cos1/x as x approaches 0 is equal to 1, which means that there is no limit as x approaches 0.

4. What are the conditions for the theorem of limit to be applied?

In order for the theorem of limit to be applied, the function must be defined and continuous in the interval around the limit point, except possibly at the limit point itself. Additionally, the limit point must be a real number or infinity.

5. How is the limit of cos1/x evaluated using the theorem of limit?

To evaluate the limit of cos1/x using the theorem of limit, we first rewrite the function as cos(1/x) and then apply the limit definition. This involves finding the limit of cos(1/x) as x approaches 0 from both the left and right sides. If the limits from both sides are equal, then the overall limit exists and is equal to that value.

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