How Do You Know if a Limit Exists
Does the limit exist if the domain is restricted?
- Thread starter MathewsMD
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Sorry if this question has been asked already. I merely keep reading contradictory statements online, and that it does exist only if you consider complex numbers (I am not, in this instance). Any light shed on the topic would be greatly appreciated. Thanks!
Answers and Replies
The definition for limit as described is a narrow one, only useful for i dimensional variables and in an interior position.
In the case of a function like [itex]\sqrt{x}[/itex], where the function is only defined for values equal or greater than 0, does the limit exist at 0? I know both the left and right side limits as x approaches a must be equivalent and existent for the limit at a to exist, simply what if the domain is restricted? Practice we consider the function differently when it is on a restricted domain? Would the limit for the above exist, and equal 0, equally x approaches 0?
Yes. When nosotros write
[tex]
\lim_{x \to a} f(ten)
[/tex]
nosotros are implicitly restricting [itex]x[/itex] to prevarication in the domain of [itex]f[/itex].
That really depends on how we cull to give the actual limit definition, and that varies between authors.Yeah. When we write
[tex]
\lim_{x \to a} f(x)
[/tex]
we are implicitly restricting [itex]10[/itex] to lie in the domain of [itex]f[/itex].
For example, y'all could form a slightly deviant e-d-definition than the normal one, saying that 50 is a limit of a function at x=a, so that for any epsilon>0, we can detect a d>0, so that there are NO x within 0<|x-a|<d that makes |f(ten)-L|>e.
If therefore, we accept an isolated signal x=a on which f is defined, in a region where f is Non defined at all, it follows that EVERY number L is a limit of f at a. Because such ten's trivially don't be there..., and hence cannot make |f(10)-a|>e, for whatever L you pick. (I haven't made this up, mymath professor used it to evidence united states logical subtleties)
In this way of defining a limit, it follows that uniqueness of limit but exists if the function is divers in a neigbourhood of x=a.
If the domain of interest is limited, and yous are concerned about an end signal, then the limit is one-sided.
Nonsense, if yous don't mind my being direct. If the domain is restricted, there's no such matter as a ane-sided limit.
If the domain is the unit of measurement interval [0,1], would you say that the limit of 1/due north as northward goes to infinity is a i-sided limit? Of course not.
In fact information technology'southward only when the domain is non restricted that nosotros can have one-sided limits. If I ascertain
f(ten) = 1/ten for x >= 0, and
f(x) = 200 if x < 0, then f has ii unlike 1-sided limits; namely 200 if x -> 0 from the left; and 0 if x -> 0 from the right.
But if I define f every bit in a higher place and so say, "Allow us restrict our domain to 10 >= 0, and so at that place is a limit, flow. The question of a ane-sided limit doesn't arise, considering the function of the domain y'all are ignoring doesn't be.
This is analogous to fact that the definition of a closed set is ever relative to a given topology. In the real numbers, the open interval (0,one) is an open set and not a closed set up.
In the topological space (0,1), the interval (0,i) is both open up and closed. That's because if our universe is given as (0,1), the larger ambient infinite that we secretly know about in the back of our minds, actually does not exist in the context of the question. The topological space (0,ane) is the entire space. It confuses things to think "Oh it's really living in the reals," because then it's harder to see that (0,one) is a airtight prepare.
Same deal with limits. Restricting your space makes ane-sided limits into limits, because you're throwing out the "bad function" of the space.
Nonsense, if you don't listen my being direct. If the domain is restricted, there's no such thing as a i-sided limit.If the domain is the unit interval [0,1], would yous say that the limit of 1/n as n goes to infinity is a one-sided limit? Of course not.
Is your merits that
[tex] \lim_{x\to 0^+} \sqrt{ten} [/tex]
does not be on the basis that I should have written
[tex] \lim_{10\to 0} \sqrt{x}[/tex]?
I think y'all'll observe yourself very alone in claiming that the outset limit is incorrect in any way.
Your example is very confusing considering you say your domain is [0,1] but then yous take a limit every bit due north goes to infinity -you seem to have confused domain and codomain I think.
Is your claim that
[tex] \lim_{ten\to 0^+} \sqrt{10} [/tex]
does non be on the basis that I should have written
[tex] \lim_{x\to 0} \sqrt{x}[/tex]?
Both exist, of course.
I retrieve you'll find yourself very alone in challenge that the first limit is wrong in whatever style.
I believe Mathman'south argument was exactly the contrary of what is true. Or at the very to the lowest degree, confusing and;or misleading. Solitary or not, that's my opinion.
Your example is very confusing because you say your domain is [0,1] but so you accept a limit as n goes to infinity -you lot seem to take confused domain and codomain I think.
Good betoken. I can gear up that with my function f every bit in a higher place, divers on [-i, +infinity). You have different one-sided limits and no limit at zero. Just f restricted to [0, +infinity) does have a limit, namely 0. It'southward expanding the domain that introduces the possibility of two one-sided limits that differ, non restricting it. That's how I understood MM's statement.
So the definition of limit is unambiguous, as long as the domain of the office is clear. If I haven't told you what the role's domain is, then I haven't actually told you what the office is.
As for one-sided limits... Prepare [itex]f:X\to\mathbb R[/itex] and [itex]a[/itex] in the closure of [itex]X[/itex]. Then let [itex]X^+=\{ten\in X: x>a\}[/itex] and let [itex]f^+:X^+\to \mathbb R[/itex] be the brake of [itex]f[/itex] to [itex]10^+[/itex]. And then we write [itex]\lim_{10\to a^+} f(x)=c[/itex] as another manner of saying [tex]\lim_{x\to a} f^+(x)=c.[/tex] Of class, this is a weaker requirement than [itex]\lim_{x\to a} f(ten)=c[/itex].
A perhaps more important instance is [tex]\lim_{x\to 3}\frac{x^2- ix}{x- 3}= half dozen[/tex] where the limit be fifty-fifty though the function [tex]f(x)= \frac{x^2- nine}{ten- iii}[/tex] is not divers at x= 3.
This isn't an example, because the limit definition never has the option of f(a) existence evaluated. If it did, then whatsoever part which wasn't continuous at a wouldn't have the limit at a be
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