My first calculus 3 limit on YouTube
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- เผยแพร่เมื่อ 17 พ.ค. 2024
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This is my first video on a multi-variable limit that you will see in your Calculus 3 class. We will evaluate the limit of y/x as (x,y) goes to (0,0) but how do we take care of this? We do get 0/0 indeterminate form but can we use L'Hopital's rule?
0:00 Limit of y/x as (x,y) goes to (0,0)
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I bought it and it was amazing 👏🏻👏🏻
What if you hat the limit as (x,y) -> (0,0) of (x^2*y)/(x^2+y^2)?
We need more of these multi-variable calculus videos. That was the math class at which I did worst in college, largely because I was burnt out after taking so many in a row. I'd like to relearn it though.
it's great! as a statistician by training and an aspiring data scientist, it's the second most useful math class out there, right behind to linear algebra. while multivariable calculus is rarely used directly nowadays unless you go into an engineering or physics-adjacent field, it's everywhere behind the scenes
Yeah, by the time I got into calculus three, my whole high school and college career had been one long string of ever-more difficult math classes, and that couldn't have been great.
There's a reason why colleges are all about those "liberal arts". General classes are good for refreshing the brain.
@@kristopherwilson506 linear algebra is the one math class I'm glad i failed. I learned a lot more the second time around and I use linear algebra all the time
Fun fact, on an intuitive level, this limit encodes every single variable limit, that approaches 0/0. So, because each of these limits have a different answer, the limit DNE.
Welcome back after century
gotta check out bprp basics!
@@kristopherwilson506 I know.
But I initially start watching this channel and than accidentally I found that there are bunch of other channel this guy has.
Now it's great to see vedio on this channel from where I started.
Please do more of these multi-variable limits
The Lamborghini Approach seems a bit luxurious, not quite straight to the point like, let's say a BlackPenRedPen lecture. But that's kind of a luxury as well, of course!
If you spend that much on a car, you want to take a longer road so you can show it off to more people of course. That's exactly the kind of route a Lamborghini would take.
@@HansVanIngelgomA curvy road enables it to show it's insane drift sounds as well
Its also easy to solve using polar coords. Substitute y=Rsin(a) and x=Rcos(a). Limit exists if and only if we get the same value for all angles "a". That's because "a" is the direction of our "attack" on the point, whereas "R" is the distance from that point. So if it does indeed matter from which side do we approach this point, then the limit at that point simply dne. Exactly the same thing as for the 1/x case, where going from 0+ yields different result than going from 0-. So after subst. we get
lim(R->0) sina/cosa = lim(R->0)
tan(a)
Which is clearly dependent on "a", therefore this limit dne. What's cool is that we can apply this exact thinking for functions with three variables., but then we must use 3D polar coordinates
This is exactly my thinking. I'd love to get @blackpenredpen 's take on the validity.
Overkill, but seems a nice tool in general
This is only necessary but not sufficient. In particular, this is the condition for f(r, theta) converging pointwise into a constant f(0, theta), but the limit (x, y) -> (0, 0) instead relies on uniform convergence of f(r, theta) into a constant f(0, theta), which is stricter.
Take a function defined in the polar coordinates with angle in [0, 2pi) as f(r, theta) = r^cotan(theta/4) if theta > 0, and 0 if theta equals 0.
If you lock the angle in place, the function r^c will approach 0 as for any c > 0 (and none with values of 0 or less are possible), which suggests a limit of 0.
However, it's also possible to take a sequence such that, for example, r = 1/2^n, theta = 4*arccotan(1/n), where it will always have a value of 1/2, even though r is approaching 0.
If this is converted back into the Cartesian coordinates, this means that for any circle around (0, 0), there is a point where it equals 1/2. Similar reasoning can find points with values 1/3, e/pi, etc. Therefore, the limit doesn't exist.
@@vladislav_sidorenko Can't we just use the radial method? Say y = mx and then substitute, so lim(x->0) of mx/x clearly depends on m, meaning that the limit does not exist.
@@venkinta3343 It's indeed a necessary condition, and it failing does mean that a limit doesn't exist, but it succeeding doesn't mean that a limit does exist (despite "if and only if" in the original comment)
My first time seeing a multivariable limit!
Please, do more of calc 3 in the videos, your explanation is awesome.
I am so glad to see this video, because my friend was trying to learn limits from me because he did not understand anything from the school teacher. I explained limits to him just like you by using graphs, and I am ecstatic to know that the graphs were not incorrect after all..
I'm grateful for your effort to spread the knowledge to us. Thank you for all this.
It’s nice to see you are finally back! ❤
So glad you're back!
Replace y with 1/2*x (Because y will still go to 0 if x goes to 0) : Get 1/2 as limit
Replace y with 1/3*x (Because y will still go to 0 if x goes to 0) : Get 1/3 as limit
Conclusion : Undefined
you choose always wonderful subjects,your method is excelent.
Thanks.
omg I’m doing that in school and you explained it so well ! Thx you
Yesss!!! You've inspired every math channel here on TH-cam!!! Thanks professor!!!🤩🤩🤩🤗🤗🤗
Hey JonnyMath!!!!
This is much simpler I think. y = rsinθ and x = rcosθ. The limit is just r going to 0, and y / x = tanθ, which doesn’t depend on r, but depends on θ so it isn’t a single value, and depends on the direction the limit is approached. Doesn’t exist
nice, more calculus 3 pls 🙏 love your content
You should do more calc 3 stuff I find it super interesting and it’s not that much harder than calc 2
THE LEGEND IS BACK
Easiest solid way: x= r cosθ, y = r sinθ, y/x= tanθ so it is not defined.
Even easier is letting y=ax (a is an arbitrary real constant). The limit becomes ax/x, meaning that the limit is equal to a. Infinite solutions, the limit does not exist.
I thought about comparing y=x to y=-x as (x,y)->(0,0) is 1 for the first but -1 for the second. It seemed natural to come at it from perpendicular lines.
its a great idea to make more videos on multivariable cauculus topics
Bro is the gamerboy80 of calculus. Always wears Nike.
no way someone made a gamerboy80 reference on a math vide LMAOO
@@anexotikbakka2342 I am happy someone got it lol
@@cdkw8254 he has ~1m subs; someone'd recognize it eventually
🗣🔥🔥🔥
amazing reference, wasnt expecting to see it here lmao
That is what we call the limit of f(x,y) in the direction of v(a,b). Notice for example that the limit of your function in the direction of (1,2) is equal to 2 and the limit in the direction of (2,1) is equal to ½ .
So only If the limits in all directions are the same, we can say that the limit of f(x,y) exists.
I havent done calc in couple years but awesome i can follow along pretty subtlely with few pauses to refresh myself. Def helped me pass my college math courses amazing channel!
Would love more multivariable limits.
Initially I thought x and y getting smaller at the same rate, so along the same diagonal from opposite directions (x=y, approaching from +ve and -ve side), so thought it was 1. Then I realised that they could go in opposite directions, so thought about the diagonal y=-x, giving -1, which means DNE. For some reason, I didn't consider that the could be getting smaller at different rates, or that one value could already be fixed at 0.
Do more Calculus 3 videos!
Definition of a limit:
Lim(f(x), x->x0) = y if and only if for all a_n with lim(a_n, n-> inf) = x0 the following statement is true: lim(f(a_n), n-> inf) = y.
This means especially if we find 2 sequences that converge to x_0 but the mapped sequence does not converge to the same value, the limit does not exist.
Easy: for f(x, y) = y/x,
just take a_n = (1/n,0) and b_n = (1/n,1/n). It is trivial to see that f(a_n) = 0 for all n and f(b_n) = 1 for all n. This means the limit does not exist.
What about changing to polar coordinates= x=r*cos(Theta), y=r*sin(Theta), then r cancels and the value of the expresion depends on Theta, aka the direction you approach (0,0), actually its tangent, which in the case x=y is one. Therefore the limit DNE
You could generally do any line y = mx. The limit will always approach m, but m can be any value meaning the limit DNE
my king has returned
I was wondering. Can I pose a point like (1/n^2 , 1/n). When n goes to infinity, Y/X = n -> +infinity. So there is no limit ?
Can you tell me if this works ?
Thank u so much for the work, it’s tremendeously good !
It would’ve been worth pointing out that the limit has a different value along _every_ line through the origin. As others have pointed out, this becomes obvious after converting to polar coordinates, an often-useful technique for similar limits.
We can tell right off the get-go that this DNE because we have no information at all on how y depends on x. If we know the relation between y and x, that is, if we know y = f(x), then we can proceed as usual. The infinite # of choices for y = f(x) corresponds directly to your infinite # of pathways to approach (0,0). Each different y=f(x) represents each of your different paths. Well done and good explanation. The power of 1 counter example!
Infinite number of pathways does not mean the limit is undefined, the infinite number of pathways can still approach the same limit. For example, lim(x,y)->(0,0) (x^2+y^2) =0. By substituting x=rcos(theta) and y=rsin(theta) and changing the limit to lim r->0 (r^2cos^2(theta)+r^2sin^2(theta)) which equals to lim r->0 r^2, we can see that the limit is independent of theta.
@@kaichingwong4358 Are you disagreeing with me or bprp? I never suggested that there not examples where the limit is path independent and therefore exists. I suppose you might try and argue that in it's present form it is indeterminate but I would disagree with that. Depending on exactly how x and y are related then the limit could exist (path independent), not exist or be an indeterminate form. Maybe x and y are not related at all which is what we have to assume here. So as it currently stands, with no other information, DNE is the right answer as was illustrated in the presentation.
@@ianfowler9340but the counterexample he gave IS path independent. And x and y are NOT related in the counterexample.
But your claim appears to be that path independence requires a relationship between x and y
@@fahrenheit2101 Am I missing something? His example of y/x gives different limits depending on the path. That makes it path dependent not independent. And that makes the limit undefined.
To your question: A relationship is a necessary but not sufficient condition for the limit to exist.
@@ianfowler9340 @ianfowler9340 Is it necessary? I don't see how the example @kaichingwong4358 gave gives different limits? x^2 + y^2? Is there a path you know of that *doesn't* give the limit 0? Do tell. If not, and they all give 0, then the limit exists, but x and y are *not* related, thus your claim of necessity isn't a real one. Indeed, you haven't proven your claim, and this appears to be a counterexample. That you keep ignoring.
thank you for helping with calc 2. is it too much to ask for you to help me pass calc 3 as well?
Great video
Discussing multiple paths is one way to show the limit does not exist, but at first blush, I thought it is simpler to use a delta-epsilon proof. If we say that there is some number delta such that for any (x, y) within some neighborhood of radius delta of some point, the value of y/x is within a neighborhood of radius delta epsilon of (0, 0), we can contradict this by pointing out that we can pick out a point on the line y = x so that y / x = 1 in any neighborhood of (x, y)
or just look at a graph
@@doktornouveau862The graph might help someone intuitively understand, but "look at the graph" does not show the answer.
@@doktornouveau862You can't really create the graph around 0 without knowledge of what happens around 0. So you might wanna check whether you actually understood the concepts.
Good job
I want more calculus 3 please
Nice video 😊❤
would you ever consider doing calc 4/vector calculus questions (im not sure if you guys have calc 4 over at the US)?
Awesome new video. Please continue bringing new content for us fans
A nice lesson.
a graphing calculator like geogebra is usually pretty handy to visualize the DNE better, tho i guess its not that clear with this one
First calc-3 limit and almost first calc-3 video in general... Please make more calc 3 videos or even better, a new Calc3 TH-cam channel please!!
Take the path y=kx and get different results for each k.
What marker do u use sir ?
Had my multivariable calculus exam yesterday and it had the same problem for f(x,y)=(x^2=2y^2)/(2x^2+y^2).
Easily my favourite maths course so far, chemical engineering student :P
Had my multivariable calculus exam yesterday and it had the same problem for f(x,y)=(x²=2y²)/(2x²+y²).
Easily my favourite maths course so far, chemical engineering student :P
Makes no sense, but at least the carets are gone.
How is your experience as a chemical engineering student? Ive been interested in it for a while
@@mrozan3578 I'm only a freshman and have yet to pick my major so it's a bit early to say but overall I've enjoyed it and there's a lot of variety in the different branches of chemical engineering you van choos from. One thing you have to be aware of is that chemival engineering includes just as much, if not more, physics as it does chemistry (mainly fluid mechanics and thermodynamics) in case that's a turnoff for you.
I found a great youtube video on the subject by Zach star, go check that out if you want to know more.
@@turkkuli3996 thanks for the recommendation on Zach Star, I've been watching his videos and it's been very helpful
Can you do a video showing a slo-mo of how you switch the pens in your hand? Or is that a Mathematicians' trade secret?
My first instinct for this problem was to reduce it into radial (r, \theta) coordinates and generalize the limit as r goes to 0. As you substitute x and y for their r cos theta, r sin theta components, the r drops out in the limit argument and it's clear that the limit is direction dependent (tan theta can take on all values...) so the limit does not exist.
can you make a video about areas of irregular shapes
awesome video! please make more calc3 content uwu
is it possible to do this via parametrization? (i think this is paramerization)
set y=tx where t is a function of x
lim x->0_ tx/x=t
since we don't know what t is the limit does not exist.
Yes, and it's not that you don't know what it is, it can actually be anything
The answer is D or Does not exist. For The limit to exist the function must be path-independent meaning regardless of which (x,y) path you take to get to the point (0,0), the function will converge on a specific value regardless of the path taken (if the limit exists) if there was a specific path in mind, then the limit would exist, but generally speaking if there is no defined path in the limit, for limits of points graphically speaking (all paths taken to reach a point, must tend towards the same value for the limit to exist) otherwise, the limit does not exist.
I think we can take a curve y=mx line where (x,y) goes to (0,0) alone this line where m is not equal to zero.
So we can say lim as (x,y) goes to zero(0,0) the limit will be m which is an arbitrary number so m can be anything. But we know that the from the uniqueness of limit we can say that for different choices of m we will get different values of the limit. So limit dne....
very nice solution
*"Bro i have one question. How to send you questions??"*
Say to me
@@madridboy3590Why did my parents get divorced?
Does lim mod(1/x)
x->0
Equals infinity?
Yes
You can show that by directly using an epsilon-delta proof. The limit to a function f at a point c is equal to +∞ if:
For all M>0 there is ε>0 such that 0
or just say that any negative value is converted into the positive value inside a modulus function, so the function mod|1/x| will just yield positive infinity@@Wakrar
Im not even a calculus 1 student as im in highschool rn and we dont have these 1 2 3 bifercation stuff.
Tho i said its dne in 3s lol.
Coz i immediately noticed there will be problem with ±values of y and x, so its ofc dne
Hello,I am in 9th standard,and I don't study calculus but this is a lot of fun to watch even though I don't understand a lot
Danny DeVito: _I get it..._
calc3 was the first (and so far only) calc course i passed
Can multivariable calculus have defined limit problems?
Please upload clac 3 study
If there a f(x,y) such that lim(x,y)->(0,0) is the same finite value on all paths except one, but on that exceptional path the limit DNE, can the limit be said to exist? Because if not, couldn't this video be made a lot shorter by saying "y/x DNE when x=0 hence it DNE overall"
The limit only exists if it exists and is equal on all paths. Who's to say that one path is wrong, and all the others are right? Maybe that one path recognised a fundamental property of the function that the others don't. So is it more important, or less?
Overall, if there's not a unilateral agreement for all possible limit types for a given problem, the root limit is the problem does not exist.
Using polar coordinates you will see that the function wont be approaching any value as the radius approaches 0.
Could you switch it into polar form to get a definite answer, or would you get the same DNE?
If you used polar cordinates y=r*sin(n) x=r*cos(n), for x and y both going to 0, it implies r goes to 0 as well, so you would get the limit equals tan(n), for any value of n, therefore it does not exist
In polar coords the function becomes rsin(θ)/rcos(θ)=tan(θ) and we'd be taking the limit as r approaches 0 from the positive direction.
This basically gives us the same situation as the value of the limit is different for different values of θ and therefore DNE.
Discover that rule brother. I would love to teach the blackpen red pen rule someday
I have a request, proof the Stirling's approximation please😢
There is a less geometric way to see this (that's how I immediately got DNE without doing any work). Without losing any generality, I can write y=f(x). So, then I have a limit of f(x)/x as x->0. Well, I do not know anything about f(x). If it is 2x then then limit is 2. 3s, then 3. There is no convergence, so DNE.
I started to make notes, but I ended up with just doodles. But I love that soft breeze in my hair, when all of this goes woosh over my head 😏
How can you do this kind of limit with delta-epsilon proof?
In theory, I have an idea why the limit DNE. Basically this limit encodes every 0/0 indeterminate form. Since each limit may have different results, this limit DNE.
Can I write this like lim x tends to y?
So nice to comment after a long time (in a new account)
How would you calculate the Limit[x!,x->-∞]?
It just occurred to me. Why isn't the limit x->0 (1/x) = ±inf ? why is it specifically DNE?
We know that specific root problems have multiple solutions so why is it not the case here?
∞ seems to mean unlimited .
-∞ means without a limit too , but negative .
when we try to unite both of them into one answer of some problem , we have already proven in words that the limit of it does not exist .
infinity and/ or unlimited , aren't they
numerous, countless, numberless .
boundless , limitless , undefined .
and i was sure it wasn’t 1 when trying to think threw it
(x, y) -> (0, 0) is not exactly the same as the behavior along the line x = y. The latter condition is stricter. You can check behavior along a different line x = 2y, (x, y) -> (0,0)
multivariable calculus is the course I had the worst grade as a math major. I have to overcome it.
Most love from India 🇮🇳 sir
Before watching the video: intuitively, I'm going to say the answer is zero? Because the 1/x part gives +/- infinity, but the y part gives 0, so whether or not the x part is + or - infinity, multiply that by 0 and we still get zero. Now, I'm sure the video will show me how wrong I am...
**casually returns after a month and a half of being dead**
gotta check out bprp basics!
we know that 'x' raised to power '0.5' = squareroot(x), but what is 'x' tetrated to '0.5'?
Why do you need to evaluate the expression first before resolving the limit once you substituted in the new values?
What do you mean? Do you mean why he simplifies the expression of the limit before letting the actual pair (x,y) tend to (0,0)? If so, because it makes it able to be evaluated.
A more basic example is
lim(x→1) (x²-1)/(x-1)
If I don't clean up the fraction and see that it's the same as x+1, how am I to comfortably solve it?
Please 🙏 Sir explain this d²y/dx²= x²sinx
What you did was show that the limit does not exist by choosing 2 very simple different paths that lead to different results. You showed the existence of the difference and the work is done.
But suppose there's a question where we genuinely don't know if the limit exists or not, and trying to choose every different path is very inefficient, is there a method you can do to solve these questions quickly?
Wassup sir where had you been 😅
Fish variable
And
Lamborghini vector
I did it in my head by choosing y=sin(x) which gives lim x->0 of sin(x) /x which equals to 1. Not sure if it is correct solution tho.
Lovely ❤❤❤🎉🎉🎉
7:50 What if x is equal to 0?
Undefined
Took this freshman year of college, got an A (I got straight A's in college), but I don't think I learned a thing.
Hello sir😊, ❤
Can you evaluate this sir.
Limx->infinity (x!/x^x) please reply sir 🙏 ❤ i couldn't find answer for this sir.😊
0
polar coordinates: x=rcosθ; y=rsinθ. So lim(y/x)=tanθ. Different θ, different limit so the answer is "does not exist"
tan θ, not cot θ
@@nightytime Ah ye
Something I either forgot or never actually knew: What does it mean for a pair to approach a particular value? Let me clarify: When we speak of f(x) having a limit as x -> a, we mean that for any epsilon > 0, there is a delta such that if |x - a| < delta, then |f(x) - L| < epsilon. One analogous definition for two variables is to say that if their Pythagorean distance is less than delta, then the f(x, y) < epsilon. But there's nothing to stop us from using |x-h| + |y-k| or some variant of these.
Just take the path y=mx
How would we prove that a limit does exist, for a different question? Because there’s no way we can check all the infinitely many possible ways to approach the target point to see that they are all equal.
If anyone can help thanks in advance
well, how do you prove that a limit exists for the single variable case? by the epsilon-delta definition, right? so probably something similar for the multivariable case.
either that or you could just "find the answer" like lim (x, y) → (1, 1) of (y² - x²)/(y - x)
One of the most popular methods for these kinds of limits is the Squeeze theorem
You just go to a different coordinate system and hope for the best. For example, polar coordinates are a good method because you can describe any possible curve with a polar function. For example, if you are approaching the origin point, you can transform the multivariable limit into a single-variable one, because the direction doesn't matter. And so your (x,y)→(0,0) changes to r → 0.
let y=kx
lim (x,y)->(0,0) y/x
lim x->0 kx/x = k
K is unknown so we have sufficient evidence to say the limit DNE.