No it's 0+(-1+/3)/2=0=-0is exdecing 0 vaule which is equilvent to nothing make it a negative vaule which us not possible so it can be a concept such as {ï 0 }
Yeah, would have made more sense to me. That second limit is sketchy, he just randomly says zero under a number that goes to zero is zero... is that true? Since both top and bottom really are limits... it would be indeterminated. Anyways
@@rodabaixo13 The thing that he did with lim 0/(2t) = lim 0 was okay because he just evaluated the term inside of the limit which is equal to 0 (for all t not equal to 0) so we just have a constant limit of zeros which is 0
Dear blackpen redpen, I found your calculus videos 6 years ago and they inspired a deep fascination in mathematics. Since then I have gone on and completed a bachelors degree in mathematics and next week I start my Masters of Research. There’s a part of me that wants thank you for putting out great entertaining math videos and inspiring the net generation of mathematicians.
faster way: let k in \doubleR set a=t b=-2kt c=tk^2, notice the sqrt cancels out so it works for the plus and the minus, the t cancels out and you get k inside the limit. that means you can get to every value you'd like under these type of linear approaches, so of course the limit does not exist, but now you can understand its beautiful graph better. if you want to look at it yourself, set x=b/a y=c/a, now you can plot it in desmos and see it for yourself :)
@@danielarnold9042 If you want a hint: Use that the rational numbers are dense in R. Use this to define the an and bn. Choose the cn such that the root is zero.
Whatever a b and c you use, as you tend them to 0 evenly it's the same as dividing both sides of the equation by the same amount. So x1 and x2 remain the same all the way to 0 where they're undefined.
But the quadratic formula expressed as f(x) would be in 4d wouldn't it? A axis B axis C axis And the f(x) axis If that is true, i dont think we could apply the definition of limit to a function in 4D, therefore we must do something similar to what you did in the path 1 and cancel some variables. Idk if when doing that the limit stays the same cuz i cant imagine how you would represent a 4D function Also probably this comment is a whole load of bs cuz im a 17 yo highschool student
Obviously this limit doesn't exist. You can change a,b,c to ka, kb, kc with k -> 0 and the formula will keep giving one number. Each real number is a root of some quadratic equation, so the set of partial limits consists of all real numbers (if we take into account complex numbers, the set of partial limits consists of all complex numbers).
It makes sense that the limit doesn't exist because the quadratic formula (without the minus before the square root) gives ONE solution to an equation (putting the abs value under the square root means there is always a solution). If all the parameters go to 0 you are basically considering a solution of the equation 0=0 which obviously has infinitely many solutions, so intuitively you get that the limit does not exist
you didnt need to compute the 2nd path, in the first path itself, the limit does not exist because LHL = (-1-sqrt 3)/2 and RHL = (-1+sqrt 3)/2 the change you made of modifying 0 to 0+ is not valid because that is not the original question
I found the solution by letter a=c=½b which made the whole formula simplify to -1, and then did a=c=b√(1/2), which made the formula (-b+√(|-b²))/(2b√(1/2)) which then simplify to (-b+b)/(2b√(1/2)) which becomes 0 at all b values, 0≠-1, therefore the limit doesn't exist
3:08 You made a mistake simplyfying t/t=1 Bc, if t=0, then we have 0/0 Right? Mmm, I think there is an other way to do it Mabye with laplace tranformation I guess
The whole point is that it's describing a solution to 0x²+0x+0=0. Every real number can satisfy that. The limit could be any real number. If you wanted the limit to exist, it would have to be a coordinate system where 0=0 isn't always true, or 0x=0 isn't always true. Good luck.
I'm in class 8, grade 8, or standard 8, so I don't know derivatives, differentials, or integrations. but I know complex numbers, some trigonometries, logarithms, and other grade 9-10 things. blackpenredpen helped me a lot to understand these. Still I don't know any calculus topic. I have invented some formulas and theorems like 3d trigonometry with 2 angles and 12 ratios, a product of some infinite sums, the relation between all types of means/averages like quintic>quartic>cubic>quadratic>arithmetic>harmonic, etc. Recently, I've been trying to make quintic formula with radicals because I don't know calculus and some special functions. many people think, it's not possible and they already have given the proof. but I think, there's still a probability that we can make quintic formula with only radicals. blackpenredpen, can you help me by explaining without calculus, why this isn't possible? please make a video on it.
And if you like means, Michael Penn did a video that shows off all the uncountably infinitely many means you can have between two numbers, in one concise formula. Okay, it contains basic calculus, but you'll pick it up quick enough. Calculus is a simple subject at heart.
@@bradleywang289 I'm also weak in graphs, and diagrams. I'm also weak in English. I have searched many videos on TH-cam about Galois theory. but all are complex. I'm waiting for blackpenredpen's reply and that video.
@@bradleywang289 I can invent a quintic formula, if and only if I solve only two symmetric equations. but the problem is, 1 of those is a 12-degree polynomial. which is super hard to solve, maybe it is impossible.
He didn't really like the dealing with negative numbers, so he chose a path (0+) that avoids any problems with negative numbers. It's really just for convenience
3:48 Can someone explain why this is allowed? Why can a=t but b & c = 0 why can a be different if the limit is still approaching 0????? (I've only taken upto Calc 2)
When working with multivariant limits, you have to consider all possible paths to get to (0,0,0). To show that the limit does not exist, we have to (for example) find 2 different paths which give a different limit. The path we consider here is t -> (t, 0, 0).
When proving lim x->0 (1/x) DNE You look at 0+ and 0- , see they disagree This is a more generalized form, with multiple variables, but the same concept (the limit should exist from all approaches)
At this point he is the quadratic formula’s worst enemy.
Bro has beef with this formula
He’s training to fight the cubic formula.
Interesting to see so many variations on quadratic formulas problem😮
You already find that limit 0+ is (-1+√3)/2 if you replace |t| with -t you find (-1-√3)/2 limit 0+ is different than 0-.so it's undefined
No it's 0+(-1+/3)/2=0=-0is exdecing 0 vaule which is equilvent to nothing make it a negative vaule which us not possible so it can be a concept such as {ï 0 }
{I 0 )
Yeah, would have made more sense to me. That second limit is sketchy, he just randomly says zero under a number that goes to zero is zero... is that true? Since both top and bottom really are limits... it would be indeterminated. Anyways
@@rodabaixo13 The thing that he did with lim 0/(2t) = lim 0 was okay because he just evaluated the term inside of the limit which is equal to 0 (for all t not equal to 0) so we just have a constant limit of zeros which is 0
@@rodabaixo13if the limit existed, it would exist from all directions; the point of this method is to choose easy directions and evaluate them quickly
Dear blackpen redpen, I found your calculus videos 6 years ago and they inspired a deep fascination in mathematics. Since then I have gone on and completed a bachelors degree in mathematics and next week I start my Masters of Research. There’s a part of me that wants thank you for putting out great entertaining math videos and inspiring the net generation of mathematicians.
under rated comment! i hope ur masters goes well
he should read his comment fr
This guy's breaking math fr
faster way: let k in \doubleR set a=t b=-2kt c=tk^2, notice the sqrt cancels out so it works for the plus and the minus, the t cancels out and you get k inside the limit. that means you can get to every value you'd like under these type of linear approaches, so of course the limit does not exist, but now you can understand its beautiful graph better. if you want to look at it yourself, set x=b/a y=c/a, now you can plot it in desmos and see it for yourself :)
That's pretty neat! 😊
The quadratic formula abuse continues...
When a=b=c=0,we will have an infinity of solutions for ax^2+bx+c,so the quadratic formula don't have an exact value in this case,so DNE
the limit is independent of the actual value. this does show that it's discontinuous at (0,0,0) though
Can you do sin(cos(x)) = cos(sin(x)) in the next video?
My teacher did this once and we found no real soln. Wonder what he will come up with 🔥
@@Player_is_I there are complex solutions
Good idea!
Shree Sridhracharya must be smoking in Mathematicians' Heaven
Bro his name was Sri Dharacharya, pls correct the comment..
Who’s that
Who discovered Quadratic Formula@@quatsch95
Shrii DharaCharya**
Oh you guys were saying for the M lol my bad
A funny exercise would be:
For any x in R, find integer sequences an, bn and cn such that the limit as n goes to infinity approaches x.
How would that even work?
@@danielarnold9042 Do you want my solution or just a hint
@@danielarnold9042 If you want a hint:
Use that the rational numbers are dense in R. Use this to define the an and bn. Choose the cn such that the root is zero.
@@johnny-wm4uo Misunderstood what you meant. Cheers.
The zeroes of the constant zero function should be everywhere, now let’s see the video :)
Well done, no limit exists.
Might also be interesting to fix a=1 or even leave it variable, then let b->=+inf and c->-inf
@@woody442 Just bt looking at it, it should be zero, right?
Yes I think so. I found one zero to be at x=0, the other is approaching -infinity
Me: What's your approach to solving quadratics?
BPRP: Tantric, I could go for hours 😩
Pls make a video on
tanx/x>1 when 0
Whatever a b and c you use, as you tend them to 0 evenly it's the same as dividing both sides of the equation by the same amount. So x1 and x2 remain the same all the way to 0 where they're undefined.
Someone's really obsessed with quadratic formula!!
It is trivial. Because when (a,b,c) =(0,0,0), clearly 0x² + 0x + 0 = 0 has no unique solution. Limit DNE.
That's interesting, can you develop a little? How is that true?
Love trivial stuff
Can you solve sumation m=0 to infinity again summation n=0 to infinity [ m!n! /(m+n+2)! Please
I thought of using l hopital rule and use quotient rule and do the calculations.
Just going to start watching this but if the result ends up being -1/12 with Rick astley playing in the background I’m gonna cry…
This man really likes quadratic formula :)
This man is gonna rest after breaking the Quadratic formula
now we DEFINETLY need a limit of the cubic formula
Can you get all solutions for a in terms of n in a=n*sqrt(n+sqrt(a))?
Depending on the way you take the limit, it can be any value in R.
But the quadratic formula expressed as f(x) would be in 4d wouldn't it?
A axis
B axis
C axis
And the f(x) axis
If that is true, i dont think we could apply the definition of limit to a function in 4D, therefore we must do something similar to what you did in the path 1 and cancel some variables. Idk if when doing that the limit stays the same cuz i cant imagine how you would represent a 4D function
Also probably this comment is a whole load of bs cuz im a 17 yo highschool student
bro will never leave the quadratic formula alone 😂😂😂
(x-a)^2=x2-2ax+a2
so an=1/n, bn=-2a/n, cn=a2/n
Q(an,bn,cn)=a and they approach zero so the limit is undefined?
Obviously this limit doesn't exist. You can change a,b,c to ka, kb, kc with k -> 0 and the formula will keep giving one number. Each real number is a root of some quadratic equation, so the set of partial limits consists of all real numbers (if we take into account complex numbers, the set of partial limits consists of all complex numbers).
Do triple intégral of quadratic formula, then do line/surface integrals with it. Abuse it more!
The quadratic formula is going to have nightmares about the black and red pen
Watching a calculus 3 video when you're just done with calculus 2 was not a good idea
Is there any proof for wavy curve method? Please make a video on it.
you cannot equate a=b=c. they could go to 0 at different rates
It’s just a path but not the only path.
Quadratic expresion =×=/b²-4ac/2ac={a,b,c}->0=lim ->0=0 quadratic expersion simplified by integration is =] lim->0
=t²=-4-t-t/2t²=0=lim->0=0
Lim->0=0 t²-4-t-t/2t²=0=lim-t->0=lim->0=0
PLEASE integrate sqrt(ax^3+bx^2+cx+d) im begging you
Can you factor the term ax^2+bx+c
K(×-r_0)(×-r_1) where k is a constemt and r_0,1 are the solutions for x of ax^2+bx+c=0. (R is complex and k is real)
Solve 3 sqrt(x)+y^(2)=512 and x^(2)+3 sqrt(y)=3125
Please solve
How do I find your email? I have a very simple questions about the course sequences before multivariables and distribution.
It makes sense that the limit doesn't exist because the quadratic formula (without the minus before the square root) gives ONE solution to an equation (putting the abs value under the square root means there is always a solution). If all the parameters go to 0 you are basically considering a solution of the equation 0=0 which obviously has infinitely many solutions, so intuitively you get that the limit does not exist
pls stop abusing the quadratic formula /j
The quadratic formula is just the solutions for x that are the roots of the polynomial in this case 0, obviously all values of x are roots.
ARE YOU DOING RESEARCH IN "Deep beneath secret of QUADRATIC FORMULA" 😅
The limit does not exist. The limit does not exist!
Will you solve one raised to omega
you didnt need to compute the 2nd path,
in the first path itself, the limit does not exist because LHL = (-1-sqrt 3)/2 and RHL = (-1+sqrt 3)/2
the change you made of modifying 0 to 0+ is not valid because that is not the original question
Where is the link for your other channel's link?
It's in the description : )
I found the solution by letter a=c=½b which made the whole formula simplify to -1, and then did a=c=b√(1/2), which made the formula (-b+√(|-b²))/(2b√(1/2)) which then simplify to (-b+b)/(2b√(1/2)) which becomes 0 at all b values, 0≠-1, therefore the limit doesn't exist
My guess is (-1+sqrt3)/2, which is about 0.366
But yes, the only formal answer is that the limit doesn't exist
3:08
You made a mistake simplyfying t/t=1
Bc, if t=0, then we have 0/0
Right?
Mmm, I think there is an other way to do it
Mabye with laplace tranformation I guess
This is so cool
I wonder if there's any set of 3d coordinates where the limit exists with the quadratic formula
Wouldn't the limit as (a,b,c) -> (1,1,-1) be (-1+√3)/2? That's more or less what happened with the a=b=c=t path in the video
The whole point is that it's describing a solution to 0x²+0x+0=0. Every real number can satisfy that. The limit could be any real number. If you wanted the limit to exist, it would have to be a coordinate system where 0=0 isn't always true, or 0x=0 isn't always true. Good luck.
You could easily show that the limit exists for (a, b, c) -> (x, y, z) if x is not 0.
However it might be a bit more work to show the other direction.
Didnt you already do this once?
It's for only as a goes to inf. This one is as all 3 variables go to 0
Interesting!
I'm in class 8, grade 8, or standard 8, so I don't know derivatives, differentials, or integrations. but I know complex numbers, some trigonometries, logarithms, and other grade 9-10 things. blackpenredpen helped me a lot to understand these. Still I don't know any calculus topic. I have invented some formulas and theorems like 3d trigonometry with 2 angles and 12 ratios, a product of some infinite sums, the relation between all types of means/averages like quintic>quartic>cubic>quadratic>arithmetic>harmonic, etc. Recently, I've been trying to make quintic formula with radicals because I don't know calculus and some special functions. many people think, it's not possible and they already have given the proof. but I think, there's still a probability that we can make quintic formula with only radicals. blackpenredpen, can you help me by explaining without calculus, why this isn't possible? please make a video on it.
Look up Galois theory, it’s not related to calculus
And if you like means, Michael Penn did a video that shows off all the uncountably infinitely many means you can have between two numbers, in one concise formula. Okay, it contains basic calculus, but you'll pick it up quick enough. Calculus is a simple subject at heart.
@@bradleywang289 I'm also weak in graphs, and diagrams. I'm also weak in English. I have searched many videos on TH-cam about Galois theory. but all are complex. I'm waiting for blackpenredpen's reply and that video.
@@bradleywang289 I can invent a quintic formula, if and only if I solve only two symmetric equations. but the problem is, 1 of those is a 12-degree polynomial. which is super hard to solve, maybe it is impossible.
Yo bprp i bet you can't solve this integral
Int(arcsin((x-2)/(x+2)))
Why do you assume that? It's not particularly hard.
x arcsin((x-2)/(x+2)) - √(8x) + 4 arctan(√(x/2)) + c
Is their a video explaining Runge kutta method plz
Errrm lim (t->0) 0/2t = 0? Not really.
I'm a med student and still watch math videos for some reason 😂
Can someone explain why he made the limit on the first path go to 0+?
He didn't really like the dealing with negative numbers, so he chose a path (0+) that avoids any problems with negative numbers. It's really just for convenience
So that the square roots remain real and simple.
Isn't this 0x^2+0x+0=0 and x can be any number
That's why the limit doesn't exist
Give this solution
"∫(e^x/ln(x))dx
3:48 Can someone explain why this is allowed? Why can a=t but b & c = 0 why can a be different if the limit is still approaching 0????? (I've only taken upto Calc 2)
When working with multivariant limits, you have to consider all possible paths to get to (0,0,0). To show that the limit does not exist, we have to (for example) find 2 different paths which give a different limit. The path we consider here is t -> (t, 0, 0).
When proving
lim x->0 (1/x) DNE
You look at 0+ and 0- , see they disagree
This is a more generalized form, with multiple variables, but the same concept (the limit should exist from all approaches)
STOP IT YOU ARE GOING TOO FAR
DNE 😐😑😑😑😑
sir pls solve the question I gave you 🥲
Which question? This is your only comment on this channel.
First❤