Does the Limit Exist if Its Not Continuous at a Point
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When does a limit not exist.
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I'm thinking its when its not continuous or differentiable.
but is there more exact definitions/am i wrong/anything really.
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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 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.A limit doesn't exist if the function is not continuous at that point.
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.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.)
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.
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.
this is only partially true...A limit doesn't exist if the function is not continuous at that point.
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...
This is not true, either.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.
[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.
A removable discontinuity (a "hole") is different from a jump discontinuity.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...
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.
Smiley faces are addicting
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
From post #3:
A limit fails to exist if the limit from the left and the limit from the right aren't equal.
Whoops my bad, I was in a hurry.
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
[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.
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