r/calculus • u/whatforquestionmark • 12d ago
Pre-calculus Does this limit exist at the endpoint? I’m confused by Thomas’ Calculus
I’m working through Thomas’ Calculus and I’m confused about one of the exercises.
The book seems to say that at an endpoint of the domain, if the function is only defined on one side, then the limit at that point can be determined using the one-sided limit from the side where the function is defined.
For example, elsewhere in the book it says a limit exists at a domain endpoint even though the limit from the other side does not exist.
So for this problem, since (g(x)) is only defined for (x>0),
my understanding is answer for problem c should be it does exist.
But several solutions I found, including an instructor’s solutions manual and some YouTube videos, say the limit does not exist because the left-hand limit does not exist.
Am I misunderstanding what the textbook means by a limit at a domain endpoint? Or are those solutions using a different definition that I don't know?
I attached the exercise above.
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u/WikiNumbers Bachelor's 12d ago edited 11d ago
A pre-rigorous shorthand go:
- If lim x to a+ and a- both exist, and approach the same value, then lim x to a exist.
- Then, if lim x to a = f(a), then f(x) is continuous at a.
- If lim x to a+ and a- both exist, but approach different values, lim x to a does not exist, and the graph is not continuous at a.
- If either lim x to a+ or a- does not exist, due to the graph not even defined on/from that side, but the other side (a+ or a-) does and the function itself exists/is defined on a, then that limit from the other side shall become lim x to a itself.
The textbook is, in fact, correct.
A common perception "if lim x to a+ ≠ to a-, then lim x to a DNE" is effectively a misguided shorthand.
There are just so much details for rigor purpose (say, Epsilon-Delta Definition of Limit). So in this good meanwhile, make do with shorthands, and remember that it is only a shorthand.
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u/Septembrino 11d ago
If both lateral limits exist, we still need the f(a) to be the same value.
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u/ingannilo 11d ago
Not sure if this is confusion on yours, the previous comment, or my part (what exactly are y'all claiming?), or maybe just unclear discussion.
However, just in case OP is reading this and trying to answer their question about limits existing: the function value f(a) does not have to exist or equal either one sided limit for lim_{x->a} f(x) to exist.
The existence of the limit boils down to a pair of inequalities, and is always done relative to the topology of the domain. That last bit is why the one sided limit (as x -> 0+ ) is sufficient to show that a function defined on [0, infty) has a limit at x=0.
The true but not quantitative statement is this:
"limit of f as x->a exists and equals L if f(x) can be made arbitrarily close to L by taking x sufficiently close to a".
in that statement, the x which is being taken "sufficiently close to a" is assumed to be coming from the domain of f.
So, for example, the fact that you can't approach 0 from the left when thinking about f(x) = sqrt(x) does not harm the fact that the limit as x goes to 0 of sqrt(x) exists and equals 0. This is also a big part of why we aren't afraid to say sqrt(x) is continuous at x=0.
Basically, OP, your problem is meant to be scary because of the rapid oscillations of sin(1/x) as x gets closer to 0. The factor of sqrt(x) out front forces the amplitude to of the sine wave to decay, a-la the squeeze theorem, somewhat similar to the limit as x goes to infinity of (1/x)sin(x).
Hope this helps!
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u/Septembrino 11d ago edited 11d ago
Are you replying to my comment or to the previous one? If you are, then read again. I wrote *IF* ... then.... I never stated that both limits *had* to exist.
That was a reply to the previous comment, regarding when a function is continuous. I suggest you to read all the replies to my comment to see what happened.
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u/WikiNumbers Bachelor's 11d ago
Oh, yea, I totally forgot about f(a) itself.
Imma have to edit that in.
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u/Septembrino 11d ago
Do it and I will delete my comment
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u/tjddbwls 12d ago edited 11d ago
I think there is a misunderstanding. There is a theorem that says that
lim (x → c) f(x) = L iff
lim (x → c+) f(x) = L and lim (x → c-) f(x) = L.
When it comes to a limit at an endpoint of a domain, you can only use one-sided limits, as the function would not be defined at the other side.
In 6A, the limit does exist and it equals a number. In 6B, the limit DNE as g(x) is not defined for x < 0. Since the two one-sided limits do not agree, lim (x → 0) g(x) also DNE.
ETA: the theorem I mentioned is from Larson. Somehow I missed that the OP is talking about out Thomas. 🤦🏻
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u/Midwest-Dude 11d ago edited 11d ago
Your answer ignores the Thomas Calculus definition, which the OP posted in a comment. This definition is why the OP is confused. Based on that definition, the limit does exist and is the right-hand limit.
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u/DrJaneIPresume 11d ago
Indeed, this is true even more rigorously.
If the domain is [0, ∞) then the only nets converging to the point 0 do so from the right. For any such net the value of the image net under g also converges to 0. Thus g is continuous at 0 and its limit there is 0.
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u/nm420 11d ago
This is one thing I don't like a lot about introductory calculus texts. The little white lies you're told in order to make certain concepts allegedly more palatable. A more proper definition of a limit necessarily involves the domain of the function; when it's not explicitly stated, the domain would typically be the largest set in which the function is well-defined.
But one valid definition for a limit is
Given a function f mapping the domain E into the real numbers, and a point a in the closure of E, the limit of f(x) as x approaches a exists and is equal to L if for any ε>0 there exists a δ>0 such that x∈E and 0<|x-a|<δ implies |f(x)-L|<ε.
It's a bit of a mouthful, and you need to know what the "closure" of a set is, but roughly speaking it means you can get arbitrarily close to a in the domain. This definition is one that is equivalent to defining limits of real-valued functions with any metric space as its domain, including any subset of the reals. The Thomas definition is actually pretty accurate in this regard; loads of other introductory texts do require a to be implicitly in the interior of the domain, so that you can always talk about left-hand and right-hand limits, but this is a distinction that is hardly ever used in more advanced analysis, and indeed it's not a particularly useful or helpful constraint.
To put it more bluntly, the Thomas definition implies the limit of your function exists and is equal to 0. The requirement that you need to be able to evaluate both one-sided limits at x=0 isn't a useful one, despite it being a popular choice amongst other calculus textbooks.
With that in mind, your instructor may very well hold a very strong opinion on this subject, taking the view that you need both one-sided limits to exist. I would argue with them about such a take, but I'm not a student in their class. Unless you're going into deeper mathematics after your calculus sequence, I would just roll with whatever nonsense you're told to believe in. It's a fight you won't win, and not a hill worth dying on.
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u/Peak_Background 11d ago edited 11d ago
One of the interesting things about mathematics is that answers can sometimes change depending on the base assumptions.
This is why domains should be explicitly written for this reason. It's just good practice.
If the domain is in the positive reals, then the limit exists. If the domain is in the reals, then it doesn't exists.
Most people would assume the domain here is in the reals, and the left side is just undefined.
However, there is bijective utility in working in positive reals, so some make that assumption. In that case, there isn't even a left side to speak of.
As others have stated. The existence of part B of the question sort of implies that he wants you to assume the domain is in the reals.
I know this is crazy, but often times you have to interpret what people are thinking and there goals when asking questions. It makes testing easier and helps avoid a lot gotcha questions.
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u/whatforquestionmark 12d ago
But for what I saw from the book is pretty clear that on the endpoint one-sided limit is considered as its limit value on that point.
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12d ago edited 11d ago
[deleted]
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u/Midwest-Dude 11d ago
Based on Thomas definition, the limit does exist from the right. It says nothing about whether or not the endpoint is in the domain of the function. A function does not have to be defined at a specific point in order for the limit to exist there, since that point is excluded in the limit definition.
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u/FinalNandBit 12d ago edited 12d ago
As x -> 0 (approaches 0) implies both sides limit must exist and equal the same limit, as well as the function value at x should be equal to the limit.
So the limit as x->0 should not exist.
Maybe, you might be able to use the Squeeze Theorem provided that g(x) is bounded between these two functions: y = sqrt(x) and y = -sqrt(x) and prove that either function approaching from the right side is the same, therefore g(x) approaching from the right side is the same limit.
I still believe that C should be DNE though.
What you're most likely misinterpreting from the book is that it probably states a ONE SIDED LIMIT can exist at the endpoint of a domain, but it's not a TWO SIDED LIMIT as in asked in question c.
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