Years ago when I first started calculus, we were taught, "As x approaches -3, f(x) approaches..." This reinforced that we weren't interested in when x was exactly -3, but interested in how f(x) behaves as we approach the value.
One way to look at it is to substitute u=x+3 and observe how 1/u^2 behaves on either side of zero. As we approach 0 from either side, 1/u^2 approaches +inf as it's always positive. The numerator is a continuous function, so its limit is the same regardless of if we approach it from + or -. Since it's negative, then the product of the two functions must be negative and thus, the limit is -inf.
6:34 when he asks you to think that 3- is a number slightly less than 3, in reality it's a number approaching (but won't equal) exactly 3 from the left, because for that matter, you can also approach a number slightly less than 3 from the right
Thanks for the graph. If the graph does not go all up or all down at the asymptote; then the answer is DNE. The typical example is y=1/x, as you go towards 0. Remember not all divide by 0 is infinity, negative infinity, or DNE. These equations are the ones that have the same root in the numerator and denominator at the same power. Those will have a hole, but otherwise be the function with the remaining values: let c(x) = (x - a)^n for some a in reals and n in integers and f(x) and g(x) be polynomials that do not have a root at a, then F = (f(x) * c(x)) / (g(x) * c(x), will be the same as f(x)/g(x) with a hole at a and the limit at a will be f(a)/g(a). Example (x+2)(x-2) / (x-2) will give you the line x+2, with a hole at x equal 2. You might see this more often as (x^2-4) / (x-2) and you have to factor it yourself.
You need to review limits. Limits are not about finding difficult values of a function. Limits of infinity are valid. As x tends to 0, 1/x tends to infinity
@@spoddieWrong, you seem to have no idea what you are talking about. As x goes to 0, the limit of 1/x does not exist. We can directly prove it by using the definition of limits. First, we need to verify there is no real valued limit. Let L be any real number. We verify, that L cannot be the limit. If L was the limit, then using the epsilon delta definition, for any epsilon > 0, there would be a delta > 0, such that for all x with 0 < |x| < delta we have |1/x - L| < epsilon. Now assuming L ≠ 0, let epsilon = |L|, then for any delta > 0, choose x = -sign(L) * delta/2. We get 0 < |x| < delta but |1/x - L| = 2/delta + |L| ≥ |L|. Assuming L = 0, let epsilon = 1 and x = min(delta/2, 1). We get |x| < delta but |1/x - L| = 1/x ≥ 1. Next, we verify that the limit is not infinity or -infinity either. If it was infinity, then for any K > 0, there would be a delta > 0 such that for any x with 0 < |x| < delta we have 1/x > K. Now just choose K = 1, and for any delta > 0 we choose x = -delta/2. We have 0 < |x| < delta but 1/x = -2/delta ≤ K. The proof for -infinity is analogous.
@@spoddie To put it a simpler and more pleasant way, 1/x tends towards infinity from the right/+ side, but tends towards negative infinity from the left/- side. This means that there is not a limit for 1/x.
@@darranrowe174 This is a Grade 10 textbook: "Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. " - Demana et al, Precalculus, 2011, p 758
Everything depends on how you define infinity! If you consider infinity without + or - to be +infinity u -infinity than everything always works and the limit exists! It is enough to unite the two definitions of limits for x approaching a definite value x0 resullting in infinity by putting an absolute value onto the f(x0)-M though! just turn it into |f(x0)-M| and the thing is fixed it's even more elegant concerning the definition because you put two definitions together into one single one (you can do this with limits going to inf with infinite value too though, resulting in just 4 definitions insread of 7)
I think of it as -3^- = -3 - delta, and -3^+ = -3 + delta, where delta is a small positive number. Thinking of it as -3.001 can be a bit more concrete.
When I took calc, I was always told If the Limit from both sides approaches Infinity, or negative Infinity the Limit doesn't exist due to unbounded behavior.
@@spoddieinfinite limits don’t exist, a limit only exists if it is equal to a finite number. Saying a limit is infinity is just an extension of notation
@@user-dh8oi2mk4fYes, they *technically* don't due to that reason, but saying it goes to infinity or -infinity is way more useful for things like improper integrals
@@user-dh8oi2mk4f It's not just an extension of notation, infinite limits exist because that's how limits are defined. Limits are not just calculating a difficult part of a function.
@@spoddie no they don’t. Infinite limits are defined to not exist, and we just say they equal +/- infinity because it’s a common case of diverging limits. Search it up on google.
Is it a bit weird to say you can always expect a 1/0 outcome to provide positive or negative infinity limit when the most trivial example of 1/x is +or- infinity at 0 depending on the sign of approach?
In this example, the sign of the approach is guaranteed to be negative, when limited to real values of x. This is what a repeated pole does, is it makes the function approach the same infinity, along both possible approach directions. When limited to real numbers, the answer is -infinity. However, when you explore all possible complex values of x, you can produce contradictory results of the limit. If you approach this from 3 + h*i, you'll get the limit as h approaches zero of 1/(h*i)^2, where h is a real number. Squaring the i, produces -1. Squaring the h, keeps it positive. Whether h is initially positive, or initially negative, 1/h^2 approaches -infinity. Multiply it with i^2, and you get +infinity. So this is an example of a contradictory result when approaching the pole from another direction, showing that the truth to the limit is DNE, when accounting for all complex numbers. It turns out, no matter how high the exponent on a repeated pole, as long as it's a finite number, there will always be at least one direction from which you can produce a contradictory limit. A 4-times repeated pole, will produce the same limit for all 4 approaches that are rectilinear with the Cartesian plane, but you'll get a contradictory limit if you approach along a 45 degree diagonal. An 8-times repeated pole, will produce the same limit for all 8 rectilinear and 45 degree diagonal directions, but you'll get your contradictory limits when approaching on 22.5 degree diagonals.
yes it was weird to say that. The way I solved the problem was by noticing that f is just (x+2)/(x+3)^2, and so x -> -3 plus or minus epsilon, you have f = -1/(plus or minus epsilon)^2 = -1 / epsilon^2 since (+1)^2 = (-1)^2 = 1. That is the true reason why you get -infinity from both directions. Was a quick 2 minute think, then I saw all this gross math he did and skipped down to the comments.
At 4:30 "By the way it's either one, never both". Really? So which one is the limit as x approaches zero of 1/x? From 6:30, you can help yourself by avoiding the phrase "less than". If you use the wording "more negative", you won't get that awkward pause at 7:13 when you've mistakenly arrived at 0+ and you're trying to rationalise what something "a little less" than -3 added to 3 sums to. I feel for you, I really do, but stick with something "more negative" than -3 and you'll intuitively arrive at 0-.
When I was in 1st year university at Waterloo starting my hon.deg. in math, our text was authored by Louis Leithold. We used to affectionately call it "the bible". Gold standard author. Gold standard textbook. "He is best known for authoring The Calculus, a classic textbook about calculus that changed the teaching methods for calculus in world high schools and universities." Pretty high praise. Here is a quote regarding: lim f(x) as x -->a = +inf ""Note: It should be stressed again that +inf is not a symbol for a real number. It can be read as "f(x) increases without bound as x --->a". In such a case the limit does not exist, but the symbol "+inf" indicates the behaviour of the values of f(x) as x gets closer and closer to a."" My professor, who was also world renowned, stressed this very thing. Case closed.
No, here it’s because no matter if you approach -3 from below or above, the denominator is positive, so the sign is given by the exponent. But this is not always true. The denominator could be always positive, negative or flip.
That is a seriously poorly worded question and the reason why I don't like multiple choice questions. Both DNE and -inf are generally OK to most instructors. Personally, I always docked a point for a solution with an infinite "limit" rather than DNE since there was a potential lack of understanding of just want a limit is with that answer. Now if the question was to describe the behavior of f(x) as x approaches -3, then something including -inf would be more appropriate. But I would NEVER EVER EVEN CONSIDER putting both DNE and -inf in the same multiple choice question.
> Now if the question was to describe the behavior of f(x) as x approaches -3, That's what a limit *is*. Now some answers are closer to correct than others, and I would be inclined to give partial credit accordingly, especially if they explained their reasoning and weren't just guessing (or blindly mashing terms together). But there's a fundamental difference between limits like this one (diverges toward -inf from both sides), or lim (x -> 0) 1/x (diverges toward +inf from one side and -inf from the other), and lim (x -> 0) 1/(2+sin(x)) (doesn't converge or diverge toward anything, instead it oscillates between 1 and 1/3 increasingly rapidly). And you can construct even weirder functions, but you get the idea.
Accepting both shouldn't be the case. They are fundamentally different, as will later be proven by either constructing a successful A-δ limit, or proving it's not even meaningful, such as by showing limsup≠liminf. One must be able to know and distinguish the two cases. lim(x→0+) 1/x=∞ lim(x→0+) sin(1/x) D.N.E.
@@nberedim "it either exists or not" Not sure that statement means what you think it means, because strictly speaking the limit does not exist. OP is correct that by definition a limit only exists if it's convergent (and converges to the same value from all directions). Limits that "equal" infinity or -infinity are divergent and so technically do not exist. So if you want to take a "it either exists or not" approach, then DNE is the only correct answer for any divergent limit. It's common practice to say that a limit "equals" infinity, simply because it tells us more about the behavior of the function beyond the fact that it's divergent, and that information is often useful, despite the fact that putting infinity after an equal sign is technically inappropriate notation. When you write that a limit = infinity, that's really short hand for "DNE because it diverges towards infinity".
I never got as far as calculus, but I do remember discussing limits in high school (Algebra 2 + Trig). This almost makes me want to try another run at math, just for fun.
Actually both DNE and -∞ are true answers. Because if a limit's answer is not a real number, then it is DNE. or a limit exists if it is converging to a real value, not diverging to ±∞. Since -∞ is not real a number (or it is a divergence) then we can say both DNE and -∞
I like your comment because it makes us think about the notation which we are using. While technically you are correct, I would not communicate the response that way. Saying that the limit in this video is DNE is similar to saying that 5+5 is in the interval (9,11). It is a true statement, but not a precise response. Indicating -infinity for this limit is preferred.
That is no longer true on complex numbers. Take delta small and positive. If we take x=i delta-3 Then (x+3)^2=-delta^2 and thus the limit become + infinity.
I'm math teacher, I say to my students: if the limit is not real, then it does not exist. Period. But if you compute it and you obtain +-infinity then you have to say: it dont exist but it goes to infinity.
Bad question. When an expression approaches + or - infinity, DNE is a correct answer. Maybe we could argue about which might be, in a sense, "more" correct or describe in a more specific way what the behaviour is close to -3. DNE is a more general statement but still correct.
Wrong. D.N.E. is only the correct answer if D.N.E. is the correct answer. This limit is well defined, and equals -∞. A limit does not exist if two directional limits give different values, if the limit supremum does not equal the limit infimum, or if the domain does not allow for every necessary direction to be considered. lim(x→∞) cos(x) D.N.E. limsup(x→∞) cos(x)=1 liminf(x→∞) cos(x)=-1 lim(x→0) 1/x D.N.E. lim(x→0+) 1/x=∞ lim(x→0-) 1/x=-∞ lim(x→0) ln(x) D.N.E. lim(x→0+) ln(x)=-∞ lim(x→0-) ln(x) cannot be taken.
@@xinpingdonohoe3978 nope, limits cannot equal infinity, the expression that's used is only a notation for a statement, it's not actually equalisation Both answers are exactly as correct as the other
@@xinpingdonohoe3978 No, a limit only exists if the function converges to the same value from every direction. If a function grows without bound towards either infinity then it is, by definition, divergent and the limit does not exist. While it's common to write that a limit 'equals' infinity or -infinity, strictly speaking this is incorrect. It gets used because it tells us the type of divergence, which is useful information despite being improper notation. It could be argued that just saying DNE is 'incomplete' as we can say more about the limit other than it being divergent, but it's not incorrect.
@@xinpingdonohoe3978 Wrong. The "equals infinity" is a nonsensical phrase. Textbook authors often use this "notation" as a stand in for "increases without bound" and clearly explain that this is an abuse of notation (an abuse of the = sign) but we will use it here in this special case as a shorthand notation. The only really correct answer is DNE. Infinity is not a destination and is not "equal" to anything.
I don't like that multiple choice question, because, although negative infinity is the best answer, it's not incorrect to say that the limit does not exist.
Brother he gave you the solution right there its -inf. Normally when you divide by 0 the reason its undefined is because we dont know wether its -inf or +inf. In this case no matter where we aproached from it would be -inf either way.
@@spoddie they're absolutely correct. The limit from the left (negative x) is -infinity. The limit from the right (positive x) is +infinity. Since there are 2 different limits the overall limit does not exist.
@@Essence1123 @zigajeglic3645 Instead of arguing about a topic you haven't studied you should go and studyi it. This is a Grade 10 textbook: "Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. " - Demana et al, Precalculus, 2011, p 758 You should ever talk about math again
@@Essence1123 @zigajeglic3645 Instead of arguing about a topic you haven't studied you should go and study it. This is a Grade 10 textbook: "Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. " - Demana et al, Precalculus, 2011, p 758 You should ever talk about math again
Looks so simple but my class couldn't figure it out:
ruclips.net/video/jECKhD4Drww/видео.html
Limits are like "it's about 'as x goes to' something, not about the something itself it's about the journey, not the destination" 🎉
Unexpected Cosmere
Years ago when I first started calculus, we were taught, "As x approaches -3, f(x) approaches..." This reinforced that we weren't interested in when x was exactly -3, but interested in how f(x) behaves as we approach the value.
Wish you did more precalc since there isnt as many good online tutors as you teaching those kind of course questions in such an eloquent way.
Thanks! Please see “bprp math basics” for that.
@@bprpcalculusbasics wait that exists 😭
One way to look at it is to substitute u=x+3 and observe how 1/u^2 behaves on either side of zero. As we approach 0 from either side, 1/u^2 approaches +inf as it's always positive. The numerator is a continuous function, so its limit is the same regardless of if we approach it from + or -. Since it's negative, then the product of the two functions must be negative and thus, the limit is -inf.
6:34 when he asks you to think that 3- is a number slightly less than 3, in reality it's a number approaching (but won't equal) exactly 3 from the left, because for that matter, you can also approach a number slightly less than 3 from the right
Thanks for the graph. If the graph does not go all up or all down at the asymptote; then the answer is DNE. The typical example is y=1/x, as you go towards 0. Remember not all divide by 0 is infinity, negative infinity, or DNE. These equations are the ones that have the same root in the numerator and denominator at the same power. Those will have a hole, but otherwise be the function with the remaining values:
let c(x) = (x - a)^n for some a in reals and n in integers and f(x) and g(x) be polynomials that do not have a root at a, then
F = (f(x) * c(x)) / (g(x) * c(x), will be the same as f(x)/g(x) with a hole at a and the limit at a will be f(a)/g(a).
Example (x+2)(x-2) / (x-2) will give you the line x+2, with a hole at x equal 2. You might see this more often as (x^2-4) / (x-2) and you have to factor it yourself.
You need to review limits. Limits are not about finding difficult values of a function. Limits of infinity are valid.
As x tends to 0, 1/x tends to infinity
@@spoddieWrong, you seem to have no idea what you are talking about. As x goes to 0, the limit of 1/x does not exist. We can directly prove it by using the definition of limits.
First, we need to verify there is no real valued limit. Let L be any real number. We verify, that L cannot be the limit. If L was the limit, then using the epsilon delta definition, for any epsilon > 0, there would be a delta > 0, such that for all x with 0 < |x| < delta we have |1/x - L| < epsilon. Now assuming L ≠ 0, let epsilon = |L|, then for any delta > 0, choose x = -sign(L) * delta/2. We get 0 < |x| < delta but |1/x - L| = 2/delta + |L| ≥ |L|. Assuming L = 0, let epsilon = 1 and x = min(delta/2, 1). We get |x| < delta but |1/x - L| = 1/x ≥ 1.
Next, we verify that the limit is not infinity or -infinity either. If it was infinity, then for any K > 0, there would be a delta > 0 such that for any x with 0 < |x| < delta we have 1/x > K. Now just choose K = 1, and for any delta > 0 we choose x = -delta/2. We have 0 < |x| < delta but 1/x = -2/delta ≤ K. The proof for -infinity is analogous.
@@spoddie To put it a simpler and more pleasant way, 1/x tends towards infinity from the right/+ side, but tends towards negative infinity from the left/- side. This means that there is not a limit for 1/x.
@@hww3136 If you're smart enough to know the eplison delta definition of limits then you should also know about infinite limits.
@@darranrowe174 This is a Grade 10 textbook:
"Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. "
- Demana et al, Precalculus, 2011, p 758
Everything depends on how you define infinity! If you consider infinity without + or - to be +infinity u -infinity than everything always works and the limit exists! It is enough to unite the two definitions of limits for x approaching a definite value x0 resullting in infinity by putting an absolute value onto the f(x0)-M though! just turn it into |f(x0)-M| and the thing is fixed it's even more elegant concerning the definition because you put two definitions together into one single one (you can do this with limits going to inf with infinite value too though, resulting in just 4 definitions insread of 7)
I think of it as -3^- = -3 - delta, and -3^+ = -3 + delta, where delta is a small positive number. Thinking of it as -3.001 can be a bit more concrete.
8:20 felt like Bewitched or I dream of Jeanie 😂
When I took calc, I was always told If the Limit from both sides approaches Infinity, or negative Infinity the Limit doesn't exist due to unbounded behavior.
I think you're remembering it Incorrectly. The function doesn't exist but the limit does.
@@spoddieinfinite limits don’t exist, a limit only exists if it is equal to a finite number. Saying a limit is infinity is just an extension of notation
@@user-dh8oi2mk4fYes, they *technically* don't due to that reason, but saying it goes to infinity or -infinity is way more useful for things like improper integrals
@@user-dh8oi2mk4f It's not just an extension of notation, infinite limits exist because that's how limits are defined. Limits are not just calculating a difficult part of a function.
@@spoddie no they don’t. Infinite limits are defined to not exist, and we just say they equal +/- infinity because it’s a common case of diverging limits. Search it up on google.
Is it a bit weird to say you can always expect a 1/0 outcome to provide positive or negative infinity limit when the most trivial example of 1/x is +or- infinity at 0 depending on the sign of approach?
In this example, the sign of the approach is guaranteed to be negative, when limited to real values of x. This is what a repeated pole does, is it makes the function approach the same infinity, along both possible approach directions. When limited to real numbers, the answer is -infinity.
However, when you explore all possible complex values of x, you can produce contradictory results of the limit. If you approach this from 3 + h*i, you'll get the limit as h approaches zero of 1/(h*i)^2, where h is a real number. Squaring the i, produces -1. Squaring the h, keeps it positive. Whether h is initially positive, or initially negative, 1/h^2 approaches -infinity. Multiply it with i^2, and you get +infinity. So this is an example of a contradictory result when approaching the pole from another direction, showing that the truth to the limit is DNE, when accounting for all complex numbers.
It turns out, no matter how high the exponent on a repeated pole, as long as it's a finite number, there will always be at least one direction from which you can produce a contradictory limit. A 4-times repeated pole, will produce the same limit for all 4 approaches that are rectilinear with the Cartesian plane, but you'll get a contradictory limit if you approach along a 45 degree diagonal. An 8-times repeated pole, will produce the same limit for all 8 rectilinear and 45 degree diagonal directions, but you'll get your contradictory limits when approaching on 22.5 degree diagonals.
yes it was weird to say that. The way I solved the problem was by noticing that f is just (x+2)/(x+3)^2, and so x -> -3 plus or minus epsilon, you have f = -1/(plus or minus epsilon)^2 = -1 / epsilon^2 since (+1)^2 = (-1)^2 = 1. That is the true reason why you get -infinity from both directions. Was a quick 2 minute think, then I saw all this gross math he did and skipped down to the comments.
I can't say why but it bothers me so much that the graph is asymmetrical
If you had a situation where you had just 0 minus, would 0 minus be negative since it is less than 0 or approaching 0 from the left?
yes
At 4:30 "By the way it's either one, never both". Really? So which one is the limit as x approaches zero of 1/x?
From 6:30, you can help yourself by avoiding the phrase "less than". If you use the wording "more negative", you won't get that awkward pause at 7:13 when you've mistakenly arrived at 0+ and you're trying to rationalise what something "a little less" than -3 added to 3 sums to. I feel for you, I really do, but stick with something "more negative" than -3 and you'll intuitively arrive at 0-.
When I was in 1st year university at Waterloo starting my hon.deg. in math, our text was authored by Louis Leithold. We used to affectionately call it "the bible". Gold standard author. Gold standard textbook. "He is best known for authoring The Calculus, a classic textbook about calculus that changed the teaching methods for calculus in world high schools and universities." Pretty high praise.
Here is a quote regarding: lim f(x) as x -->a = +inf
""Note: It should be stressed again that +inf is not a symbol for a real number. It can be read as "f(x) increases without bound as x --->a". In such a case the limit does not exist, but the symbol "+inf" indicates the behaviour of the values of f(x) as x gets closer and closer to a.""
My professor, who was also world renowned, stressed this very thing. Case closed.
I don't get it there isn't a limit because it doesn't approach a number ±inf isn't a number so the limit doesn't exist
So if you have 0 in the denominator then the exponent tells you the sign of the infinity.
No, here it’s because no matter if you approach -3 from below or above, the denominator is positive, so the sign is given by the exponent. But this is not always true. The denominator could be always positive, negative or flip.
Not exactly.
-1/x and 1/x have the same exponents, but differing coefficients. From each direction, the limits are different.
That is a seriously poorly worded question and the reason why I don't like multiple choice questions. Both DNE and -inf are generally OK to most instructors. Personally, I always docked a point for a solution with an infinite "limit" rather than DNE since there was a potential lack of understanding of just want a limit is with that answer. Now if the question was to describe the behavior of f(x) as x approaches -3, then something including -inf would be more appropriate. But I would NEVER EVER EVEN CONSIDER putting both DNE and -inf in the same multiple choice question.
> Now if the question was to describe the behavior of f(x) as x approaches -3,
That's what a limit *is*. Now some answers are closer to correct than others, and I would be inclined to give partial credit accordingly, especially if they explained their reasoning and weren't just guessing (or blindly mashing terms together). But there's a fundamental difference between limits like this one (diverges toward -inf from both sides), or lim (x -> 0) 1/x (diverges toward +inf from one side and -inf from the other), and lim (x -> 0) 1/(2+sin(x)) (doesn't converge or diverge toward anything, instead it oscillates between 1 and 1/3 increasingly rapidly). And you can construct even weirder functions, but you get the idea.
Accepting both shouldn't be the case. They are fundamentally different, as will later be proven by either constructing a successful A-δ limit, or proving it's not even meaningful, such as by showing limsup≠liminf. One must be able to know and distinguish the two cases.
lim(x→0+) 1/x=∞
lim(x→0+) sin(1/x) D.N.E.
The question IS CLEARLY about the limit (lim). It's not a just-made-it- up symbol and definition. It's very rigorous, and it either exists or not.
@@nberedim "it either exists or not"
Not sure that statement means what you think it means, because strictly speaking the limit does not exist. OP is correct that by definition a limit only exists if it's convergent (and converges to the same value from all directions). Limits that "equal" infinity or -infinity are divergent and so technically do not exist. So if you want to take a "it either exists or not" approach, then DNE is the only correct answer for any divergent limit.
It's common practice to say that a limit "equals" infinity, simply because it tells us more about the behavior of the function beyond the fact that it's divergent, and that information is often useful, despite the fact that putting infinity after an equal sign is technically inappropriate notation. When you write that a limit = infinity, that's really short hand for "DNE because it diverges towards infinity".
Fail. Back to school!
I never got as far as calculus, but I do remember discussing limits in high school (Algebra 2 + Trig). This almost makes me want to try another run at math, just for fun.
I thought LHopital’s rule would work here.
(tanx+secx-1)/(tanx-secx+1) =secx+tanx
first one i could solve only by glacing at it
Actually both DNE and -∞ are true answers. Because if a limit's answer is not a real number, then it is DNE. or a limit exists if it is converging to a real value, not diverging to ±∞. Since -∞ is not real a number (or it is a divergence) then we can say both DNE and -∞
I like your comment because it makes us think about the notation which we are using. While technically you are correct, I would not communicate the response that way. Saying that the limit in this video is DNE is similar to saying that 5+5 is in the interval (9,11). It is a true statement, but not a precise response. Indicating -infinity for this limit is preferred.
You need to review limits
@@spoddie This limit is -infinity when limited to the real numbers, but in general, it's DNE.
That is no longer true on complex numbers. Take delta small and positive. If we take x=i delta-3 Then (x+3)^2=-delta^2 and thus the limit become + infinity.
Sketch a graph.
Watch the video
The expression does not converge to a real number, so saying the limit DNE is correct.
bro forgot what a limit was
I'm math teacher, I say to my students: if the limit is not real, then it does not exist. Period. But if you compute it and you obtain +-infinity then you have to say: it dont exist but it goes to infinity.
No L'hopital?
no bcz on plugging the limits, it doesn't give you any indeterminate form
L'Hôpital's is for indeterminate forms, specifically 0/0 and ∞/∞. This fraction is neither 0/0 or ∞/∞.
you are right. thanks
Bad question. When an expression approaches + or - infinity, DNE is a correct answer. Maybe we could argue about which might be, in a sense, "more" correct or describe in a more specific way what the behaviour is close to -3. DNE is a more general statement but still correct.
Wrong. D.N.E. is only the correct answer if D.N.E. is the correct answer.
This limit is well defined, and equals -∞.
A limit does not exist if two directional limits give different values, if the limit supremum does not equal the limit infimum, or if the domain does not allow for every necessary direction to be considered.
lim(x→∞) cos(x) D.N.E.
limsup(x→∞) cos(x)=1
liminf(x→∞) cos(x)=-1
lim(x→0) 1/x D.N.E.
lim(x→0+) 1/x=∞
lim(x→0-) 1/x=-∞
lim(x→0) ln(x) D.N.E.
lim(x→0+) ln(x)=-∞
lim(x→0-) ln(x) cannot be taken.
@@xinpingdonohoe3978 nope, limits cannot equal infinity, the expression that's used is only a notation for a statement, it's not actually equalisation
Both answers are exactly as correct as the other
@@xinpingdonohoe3978 No, a limit only exists if the function converges to the same value from every direction. If a function grows without bound towards either infinity then it is, by definition, divergent and the limit does not exist.
While it's common to write that a limit 'equals' infinity or -infinity, strictly speaking this is incorrect. It gets used because it tells us the type of divergence, which is useful information despite being improper notation. It could be argued that just saying DNE is 'incomplete' as we can say more about the limit other than it being divergent, but it's not incorrect.
@@aaykat6078 for a real function of a real variable, limits are allowed to take extended real values, and are allowed to not exist.
@@xinpingdonohoe3978 Wrong. The "equals infinity" is a nonsensical phrase. Textbook authors often use this "notation" as a stand in for "increases without bound" and clearly explain that this is an abuse of notation (an abuse of the = sign) but we will use it here in this special case as a shorthand notation. The only really correct answer is DNE. Infinity is not a destination and is not "equal" to anything.
I don't like that multiple choice question, because, although negative infinity is the best answer, it's not incorrect to say that the limit does not exist.
Brother he gave you the solution right there its -inf. Normally when you divide by 0 the reason its undefined is because we dont know wether its -inf or +inf. In this case no matter where we aproached from it would be -inf either way.
@@Paketaqiyeah but even if a limit can be shown as infinity it still technically doesn’t exist
@@thepuglover5450It does. The answer is getting closer and closer to neg infinity as you approach-3.
Which is what a limit is.
@@thepuglover5450This is not true.
@@thepuglover5450what? No! Who told you that?
Seems like a bunch of commenters need to learn about limits. Eg
As x tends to 0, 1/x tends to infinity
depends on the side you are approaching 0 from so the lim does not exist
@@zigajeglic3645 you should have learnt about that in junior high school.
@@spoddie they're absolutely correct. The limit from the left (negative x) is -infinity. The limit from the right (positive x) is +infinity. Since there are 2 different limits the overall limit does not exist.
@@Essence1123 @zigajeglic3645
Instead of arguing about a topic you haven't studied you should go and studyi it. This is a Grade 10 textbook:
"Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. "
- Demana et al, Precalculus, 2011, p 758
You should ever talk about math again
@@Essence1123 @zigajeglic3645
Instead of arguing about a topic you haven't studied you should go and study it. This is a Grade 10 textbook:
"Sometimes the values of a function of can approach different values from the approaches a number c from opposite sides. When this happens, the limit of f as x approaches c from the left is the left-hand limit of f at x and the limit of f as x approaches c from the right is the right-hand limit of f at c. "
- Demana et al, Precalculus, 2011, p 758
You should ever talk about math again