Prove Limit Existence: sin(1/x) as x→0

In summary, a good method to prove whether or not a limit exists is to use the definition of limit, which states that for every 'e' greater than 0, there exists a 'd' greater than 0 such that for |x-x0| < d, |f(x)-L| < e. For practical purposes, one can evaluate the right-sided and left-sided limits at the point in question. If the limit does not exist at that point, the definition of limit will not hold. For example, the limit of sin(1/x) as x approaches zero does not exist because the function 1/x has a discontinuity at x=0 and the definition of limit does not hold for that point.
  • #1
aquitaine
30
9
In general, what is a good method to prove whether or not a limit exists? For example limit of sin(1/x) as x approaches zero.
 
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  • #2
Hi, ill recommend you to take a look to the definition of limit:
For every 'e'>0,exists 'd'>0, such that for /x-x0/<d then
/f(x)-L/<e
If that holds, then the function f(x) has limit at the point x0,
and it is equal to L.
Of course a bold statement like this doesn't seem to help to much, so here is what you do for practical purpouses:
the function 1/x has a discontinuity at x=0, so evaluate the
rightsided and leftsided limits at x=0:
lim x->0-...1/x=-inf, and lim x->0+...1/x=+inf.
Then the definition of limit doesn't hold for x0=0 and therefor the limit doesn exists at this point.
Hope i helped you.
Have a good day.


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FAQ: Prove Limit Existence: sin(1/x) as x→0

What is the definition of a limit?

A limit is the value that a function approaches as the input of the function approaches a certain value or point.

How do you prove the existence of a limit?

To prove the existence of a limit, we must show that the function approaches the same value regardless of which direction the input is approaching the point in question.

Why is the limit of sin(1/x) as x→0 undefined?

The limit of sin(1/x) as x→0 is undefined because as x approaches 0, the value of 1/x approaches infinity and the sine function oscillates infinitely between -1 and 1. Therefore, the limit does not converge to a single value.

Can the limit of sin(1/x) as x→0 be proven using the squeeze theorem?

Yes, the squeeze theorem can be used to prove the limit of sin(1/x) as x→0. By bounding the function between two simpler functions whose limits can be easily calculated, we can show that the limit of sin(1/x) must also exist and be equal to the limits of the bounding functions.

Are there any other methods to prove the existence of the limit of sin(1/x) as x→0?

Yes, the limit can also be proven using the definition of a limit, by showing that for any given epsilon value, there exists a corresponding delta value such that the distance between the function and the limit is less than epsilon whenever the distance between the input and the point in question is less than delta.

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