When does a limit not exist.

  • Thread starter brandy
  • Start date
  • #1
161
0

Main Question or Discussion Point

I cant remember exactly.
I'm thinking its when its not continuous or differentiable.
but is there more exact definitions/am i wrong/anything really.
 
Last edited by a moderator:

Answers and Replies

  • #2
274
1
A limit doesn't exist if the function is not continuous at that point.

The way to find out if a limit of a certain function exists or not is to approach the limit from the left and the right side.

For example: Take the limit of the function f(x) as x approachs 0. If you approach 0 from the left and it equals -inf and when you approach 0 from the right and it equals inf then the limit of f(x) as x approachs 0 doesn't exist.

In this case it doesn't exist because it is infinite discontinous. (Not sure if thats the right term to use. Please correct me if I'm wrong.)
 
  • #3
33,634
5,294
A limit doesn't exist if the function is not continuous at that point.
That's not true. As a counter example, consider f(x) = (x2 - 9)/(x - 3). This function is discontinuous at x = 3, yet its limit as x approaches 3 is 6. The graph of f is identical to the graph of y = x + 3 except that the graph of f has a discontinuity (a hole) at the point (3, 6). This kind of discontinuity is called a removable discontinuity.
The way to find out if a limit of a certain function exists or not is to approach the limit from the left and the right side.

For example: Take the limit of the function f(x) as x approachs 0. If you approach 0 from the left and it equals -inf and when you approach 0 from the right and it equals inf then the limit of f(x) as x approachs 0 doesn't exist.

In this case it doesn't exist because it is infinite discontinous. (Not sure if thats the right term to use. Please correct me if I'm wrong.)
That's the basic idea. A limit fails to exist if the limit from the left and the limit from the right aren't equal.

A function can be unbounded and we say its limit is infinity if the left- and right-side limits are the same. For example,

[tex]\lim_{x \to 0} \frac{1}{x^2} = \infty[/tex]

In one sense, a limit doesn't exist, since infinity is not a number. The closer x gets to 0 on either side, the larger 1/x2 gets. However, the left- and right-side limits are both doing the same thing.
 
  • #4
274
1
That's not true. As a counter example, consider f(x) = (x2 - 9)/(x - 3). This function is discontinuous at x = 3, yet its limit as x approaches 3 is 6. The graph of f is identical to the graph of y = x + 3 except that the graph of f has a discontinuity (a hole) at the point (3, 6). This kind of discontinuity is called a removable discontinuity.
That's the basic idea. A limit fails to exist if the limit from the left and the limit from the right aren't equal.

A function can be unbounded and we say its limit is infinity if the left- and right-side limits are the same. For example,

[tex]\lim_{x \to 0} \frac{1}{x^2} = \infty[/tex]

In one sense, a limit doesn't exist, since infinity is not a number. The closer x gets to 0 on either side, the larger 1/x2 gets. However, the left- and right-side limits are both doing the same thing.
Thanks for clarifying.
 
  • #5
121
0
A limit doesn't exist if the function is not continuous at that point.
this is only partially true...


Mark44 pointed out this exception:
That's not true. As a counter example, consider f(x) = (x2 - 9)/(x - 3). This function is discontinuous at x = 3, yet its limit as x approaches 3 is 6. The graph of f is identical to the graph of y = x + 3 except that the graph of f has a discontinuity (a hole) at the point (3, 6). This kind of discontinuity is called a removable discontinuity.
.


...and here is another: find the limit of f(x) = |x| as x approaches 0.

its graph looks like f(x) = x for all x > 0, and f(x) = -x for all x < 0. notice that, despite the fact that there are no discontinuities in the graph of this function, the limit of f(x) = |x| as x approaches 0 from the right is negative 1, while the limit of f(x) = |x| as x approaches 0 from the left is positive 1. therefore the limit of this function does not exist at x = 0. don't be fooled into thinking that you can always calculate the limit anywhere along a function just b/c it is continuous everywhere. the scenario Mark44 showed above is called a removable discontinuity, or a jump discontinuity. there is a name for the discontinuity i described (where the graph of a function comes to a sharp point somewhere, but remains continuous), i just can't think of the name of it right now...
 
  • #6
27
0
A limit does not exsist when a different limit is aproached from the right then from the left. Situations like this occur frequently in piece wise functions.
 
  • #7
33,634
5,294
Mark44 pointed out this exception:

...and here is another: find the limit of f(x) = |x| as x approaches 0.

its graph looks like f(x) = x for all x > 0, and f(x) = -x for all x < 0. notice that, despite the fact that there are no discontinuities in the graph of this function, the limit of f(x) = |x| as x approaches 0 from the right is negative 1, while the limit of f(x) = |x| as x approaches 0 from the left is positive 1.
This is not true, either.
[tex]\lim_{x \to 0} |x| = 0[/tex]

The absolute value function, f(x) = |x|, is continuous everywhere.

You seem to be thinking about the derivative of this function.

therefore the limit of this function does not exist at x = 0. don't be fooled into thinking that you can always calculate the limit anywhere along a function just b/c it is continuous everywhere. the scenario Mark44 showed above is called a removable discontinuity, or a jump discontinuity.
A removable discontinuity (a "hole") is different from a jump discontinuity.
there is a name for the discontinuity i described (where the graph of a function comes to a sharp point somewhere, but remains continuous), i just can't think of the name of it right now...
 
  • #8
121
0
This is not true, either.
[tex]\lim_{x \to 0} |x| = 0[/tex]

The absolute value function, f(x) = |x|, is continuous everywhere.

You seem to be thinking about the derivative of this function.
oops, you're absolutely right - i was thinking of the derivative of |x|, which = -1 for all x < 0, and +1 for all x > 0. must be my mid-day brain fart lol. i see clearly now that, despite the fact that the graph of f(x) = |x| is not smooth at x = 0, x does approach 0 from both sides...my mistake.
 
  • #9
371
1
a limit doesn't exit when the behaviour of a function approaching a value remains ambiguous. Think about it: the whole idea of a limit is to characterize the behaviour of a function as you are close to a value you wish to inspect. Now, to think of when this cannot be discussed (i.e., when a limit does not exist), we have two choices: either the function is not defined there at all ( this is trivial when talking about whether or not a limit exists ), or the behaviour is ambiguous ( left and right limits are not the same )
 
  • #10
353
1
I don't know if anyone said this, but also when the right hand and left hand limits are not the same.


Smiley faces are addicting:devil:
 
  • #11
33,634
5,294
I don't know if anyone said this, but also when the right hand and left hand limits are not the same.


Smiley faces are addicting:devil:
From post #3:
Mark44 said:
A limit fails to exist if the limit from the left and the limit from the right aren't equal.
 
  • #13
1,101
3
Well there is the negation of the epsilon-delta definition, but in practice you sometimes use the entire range of calculus tools to determine whether a certain limit does or does not exist.
 
  • #14
161
0
so it doesnt matter if it is able to be differentiated?




i asked this question because im very rusty on my basic maths and i want to mathematically determine if a peicewise function is continuous and differentiable at the point the function changes. so i said if a limit exists it is both. because i seem to recal that a function needs to be both differentiable and continuous for a limit to exist. perhaps i am wrong :S
 
Last edited:
  • #15
HallsofIvy
Science Advisor
Homework Helper
41,833
955
In order that a function be differentiable at a point, it must be continuous which means that the limit, [itex]\lim_{x\to a}f(x)[/itex] exist. It is also required that the limit of the "difference quotient,
[tex]\lim_{h\to 0}\frac{f(a+h)- f(a)}{h}[/tex]
exist.

If is possible that a function be continuous at a point yet not be differentiable there. The example |x|, given earlier, is continuous at x= 0 but not differentiable there.
 

Related Threads on When does a limit not exist.

Replies
4
Views
3K
  • Last Post
Replies
11
Views
5K
Replies
4
Views
2K
Replies
2
Views
7K
Replies
3
Views
708
Replies
12
Views
3K
Replies
10
Views
17K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
683
  • Last Post
Replies
12
Views
5K
Top