(infinity-infinity)^infinity
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- čas přidán 29. 11. 2017
- Have you ever seen an indeterminate form INSIDE of an indeterminate form for a calculus limit question? I came up with this very indeterminate calculus limit for my calculus 1 students! Here we will see how to use the square root conjugate, taking the logarithm, and the L'Hôpital's Rule to evaluate this very indeterminate limit challenge! #calculus #limit #blackpenredpen
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This is the vilest limit I've ever seen and if I ever become a calculus teacher I'll definitely put it in one of the exams
Kiritsu u evil beast
Just make it the exam. Or the extra credit that gives you a huge boost.
While I have to admit it was quite tiring and time consuming, methodology wise, it was pretty simple. Not even comparable to some monster like lim( n!/n^n)^(1/n) with n approaching infinity. That’s by far the most difficult one I have ever seen
@@giannispolychronopoulos2680 yea that one is definition of hard question-simple answer.
not good not good we need more proofs related questions
To be honest, if I got that in an exam, about half-way through working it out I'd assume that I'd done something wrong and just give up and move on to the next question.
I mean the exam would have to be just that question because you're using every rule in it anyway haha
Ben McKenzie This solution is unnecessary complicated, with all the fraction multiplying.
Just collect x, do a MacLaurin expansion of the square roots, stopping at the first term after the 1 in order not to get 1^inf. It's going to be:
lim (1 + a/x)^x = e^a, with a = -1/2,
which you should know, or if you don't, you can easily find with De l'Hopital.
landochabod7 I have no idea what you’re talking about, and this is from somebody who understands everything bprp’s talking about!
I thought the same.
@@tcocaine it would be best to give this as a homework exercise.
If you look up the word "tedious" in the dictionary, you will almost certainly find a picture of this limit problem.
John Dixon
If you search calc2 final exam, then... :)
This is an easy one. Next up, find the limit of the needed angles to perform 3 consecutive orthographic rotations in Polar Coordinates using Quaternions while mapping it to a 3D Complex Cartesian plane where those rotations do not change its original orientation!
@@kausarmeutuwah8304 Haha! I was being a bit sarcastic! That type of problem with all of the multivariable unknowns and many partial derivatives and multiple integrations would be insane! It would probably take 3-5 whiteboards and about 30 pens and about 15 hours to do it by hand! Let's just open up Wolfram or Mathlab... let it do it for us! Get the response, can not compute!
Heavyyyyyy!!!!
Heavyyyyyy!!!!
The teachers I have had in the past would hate that answer. They'd want to get the square roots out of the denominator, and have (square root e)/e instead.
Perhaps. In Calculus, it's customary to leave the result unrationalized.
after doing a question like that i wouldnt give a damn about rationalising the fraction lol
You mean e^1/2 ?
@@Nonexistility They just rationalized it
That's the difference between an algebra teacher and a calculus teacher.
Bursts out laughing when you said pray your calculus 2 teacher doesn’t see this. Thank goodness I already finished calc 2
nadia salem
(I am actually a calc2 teacher loll)
blackpenredpen I just hope you don’t give this one on the finale unless it’s the only question
It's been 6 years since I took calc 2, although my teachers covered this type of limit in calc 1.
Square root of e... That's when you know you broke mathematics.
Or when you find out that a subset has more elements than the original set
@@seroujghazarian6343 That's not hard to do. How many integers are there in the set of values in this inclusive range of [0,1]? Simply 2. Now, how many reals are there in that same inclusive range? Infinite!
@@skilz8098 Yeah, but considering ]0,1[ has as many elements as R (cot(πx) being the bijection between them)...
@@seroujghazarian6343 Nice!
I like my conjecture...
Every single concept of mathematics, all branches, and levels are all embedded in the simple expression (1+1) ...
Everything is derived or integrated from it. Just the act of adding 1 to itself, the application of applying the operation of addition which is a linear transformation, translation to be exact defines the unit circle. There is perfect symmetry, reflection, and a 180 degree or PI radians rotation embedded within it.
It isn't directly obvious at first, but take a piece of paper and mark a point on it and draw a line segment of an arbitrary distance towards your right. Now label the starting point 0 and the ending point 1. To add 1 to this line segment or unit vector is to take the total length or its magnitude and translate it along the same line in the same direction. The tail of the new vector will be at the head of the original and the head of the new vector will be pointing at 2.
By doing this the total distance is from the initial point of 0 to the new location will be 2. This turns the expression of (1+1) into the equation (1+1) = 2. This equation is actually the definition of both the Pythagorean Theorem and the Equation of the Unit Circle that is positioned at the origin (0,0).
You see, we went to the right from the starting point and labeled that 1. We could of went to the left and labeled it 1 as well. However, they are opposing directions, and vectors have two parts, magnitude its length, and its sign or direction or angle of rotation. So, we can label the point to the left with -1.
Here the starting point of 0 is the point of reflection, point of symmetry and the point of rotation. If we rotate the point 1 to the point -1 with respect to the initial point 0, you will make an arc that is PI radians which also takes you from 1D space into 2D space. Now we have the Y coordinates as well.
This is simply due to 1+1 = 2 which is also 1x2 = 2. And this is evident because we know that a circle with a radius of one has a diameter of 2, its Circumference is 2*PI and its Area is PI units^2.
We know that the Pythagorean Theorem is C^2 = A^2 + B^2. We also know that the equation of a circle centered at the origin is r^2 = x^2 + y^2... They are the same exact equation.
When we look at the general equation of a line in the form of y = mx+b we know that m is the slope between two points and b is the y-intercept. The slope is m = (y2 - y1)/(x2-x1).
We can let m = 1, and b = 0 and this gives us y = x. A diagonal line that goes through the origin. This line has an angle of 45 degrees or PI/4 radians above the X-axis.
We know by definition that the slope is rise/run. We also see that it is (y2-y1)/(x2-x1) which is also dy/dx, change in y over change in x. If we look closer we can see that dy/dx is also sin(t)/cos(t) where (t) is the angle above the x-axis. This is also tan(t).
When you look at the original equation of the line y = mx+b we can see this as f(x) = tan(t)x + b. This all comes from 1+1 = 2!
Every polynomial, every geometrical shape, including vectors and matrices, their operations, even concepts in calculus such as derivatives and integrals are rooted in (1+1)!
I just love how everything within mathematics is all connected and related! I'm sure you know all of these concepts but was just wanting to illustrate all of their interconnections.
Yes, I take a Physical approach to mathematical induction! Why? Because without physics, or the ability to move, or translate, then the operation of addition would have no application or meaning! You can not even add 1 to itself in a scalar manner without treating them first as vector quantities. The number 1 itself is the unit vector, and the unit vector is the number 1 itself. You need physics, motion to perform the operation of addition! And as soon as you have motion, you have, limits, derivatives, and integrals!
skilz8098 very well written!
imagine you want to show this in a perfect formal way. I think the video would go over 1 hour. But a real good problem in which you can check if you really confident in limits.
namensindüberbewertet yea I know. That's why I just circled circled loll
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : czcams.com/video/3vAvpeqJkEs/video.html
@@pullingrabbitsouttaahat I would see it but you spelled l'Hôpital wrong
@@JakubS Thanks For Pointing Out. But I Don't Care Much Useless Things.
As soon as I saw it, I knew it would lead to 1^infinity
What I was not ready for was the bit after. I love your videos and they've genuinely helped me.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : czcams.com/video/3vAvpeqJkEs/video.html
@@pullingrabbitsouttaahat this link is probably an add to his content that seems controversial, probably not something to waste time one except if you wanna laugh.
This is honestly the best limit I have ever seen, it has literally everything a student needs to be able to calculate limits (when x goes to infinity). Absolutely amazing.
I just failed my calculus test, and after having watched so many videos on youtube on how to solve indeterminates, the algorithm has recommended me this.
Never before have I been so happy to never have stumbled upon this equation. I'm pretty sure that it would have broken me if it had appeared in the test. Now I have newfound respect towards normal indeterminates. This right here is the eldritch equivalent of indeterminates. It makes me shudder to think how you even stumbled upon it.
This is those questions that you skip without even looking at it
The evil Laugh at 17:18 x)
: )
I am really happy to see 20 minutes video. Yay indeed! Thanks!
Crazy Drummer I am very glad too. Thank you!!!
This is truly great maths to watch, and one hell of a limit problem. Thanks for posting.
Lies again? DMP Triple
This idea of ignoring lower order terms reminds me of something we call O-notation in rough algorithmic time and memory measurement calculations.
You can do this problem with a Taylor approach, which ends up being formally written with oh notation (you need your error in the brackets to be o(1/x) to get the right final answer).
I just finished my calc I final exam and I gotta say I’m pretty excited for calc II. Thank you for such a great rigorous video to end my night :D
congrats on finishing calc 1
What a patient guy...
I usually edit a step to change it into the next step and this way I save a lot of ink and space. But he is so hard working. Hats Off..
Am I the only one that got cracked up when he said "oh, I drew the same box again" I spent 5 minutes plus laughing man
If we substitute x with 1/y , y approaching 0 in the positive area we can jump a lot of algebric steps.
I watch it all
amazing and very complicated differentiation work
thank you
Best Xmas gift ever!
upside-down thumbnail
This is more of a hand workout than a calculus question 😂😂😂
Amazing Video! Had a fun time watching because you made it easy to follow
that was süper brütal
also, as a christmas gift almost as good as socks.
AndDiracisHisProphet hahahahahah. I like those dots on the u
that's the second time derivative.
Wats wrong with socks? I could use a few pairs. :)
It's like gifting your girlfriend with a vacuum cleaner or an iron
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : czcams.com/video/3vAvpeqJkEs/video.html
A great feat of perseverance using various interesting techniques. Awesome 👌
I'm glad that you're not a lecturer in my univ, or else the exam will be hellish haha
Max Haibara tbh I think this problem would be fun but it would take me hours at least
I just want to know who drew the house that you can faintly see on the left side of the whiteboard
Maths will be interesting if you are my teacher. I remembered that today is teachers day.
Happy teachers day
Wow, that was really crazy! Great video!
An awesome limit I've ever seen. Love it.
Yo Dawg, I heard you like indeterminate forms, so I put an indeterminate form on your indeterminate form so you can calculate limits while you calculate limits!
one of the most insane limits ever seen! a huge madness..
By doing a little more algebra on the first limit calculation (that resulted in 1) you get 2x/(2x+1) (which is 1 - 1/(2x+1) ). So the original limit is (2x/(2x+1))^x. By substituting m=2x+1, this results in (1-1/m)^(m/2 - 1/2). Using lim as n->inf (1+a/n)^bn = e^ab, the limit is e^(-1/2).
This is the best answer
Beautiful proof! Great job
Enjoyed that :)
This 22 mins was the best part of my day...
Oh... my... God. That's some seriously insane algebra. Talk about a test of knowledge of detail! Well done.
e comes up when you do not expect it. Brilliant!
Well you nailed it at the end, a lot of fun watching it.
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : czcams.com/video/3vAvpeqJkEs/video.html
I've watched so many blackpenredpen videos enough to start understanding what it is I'm watching, I am so happy ~
Sqrt(e)/e makes my calc teacher happier. Great work amazing video!
could you expand the 2 square roots using the binomial theorem, get some cancellations when taking the difference, and somehow put a cap on the remainder?
I saw this video on my recommended today and even though it is late I just had to comment on what I found.
Changing the constants by 1 multiplies the solution by a factor of 1/√e
As in lim(x->inf) of (√x²+2x+4-√x²+3)^x = 1 and lim(x->inf) of (√x²+2x+2-√x²+3)^x = 1/e
The other constant also works similarly lim(x->inf) of (√x²+2x+3-√x²+2)^x = 1 and lim(x->inf) of (√x²+2x+3-√x²+4)^x = 1/e
Basically the solution comes out as lim(x->inf) of (√x²+2x+a-√x²+b)^x = e^[½(a-b-1)]
I would never have guessed that just by looking at the equation.
Not sure if this makes it any simpler but since you have f(x)^x as a limit, maybe you could use the limit that defines e^x? Or in this case, e^k? Like so...
[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3)]^x = [1 + k/x]^x
solving for k gives
k = x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1]
but [1 + k/x]^x is just e^k in the limit
you still have to work out that x*[sqrt(x^2 + 2x + 3) - sqrt(x^2 + 3) - 1] is equal to -1/2 in the limit
I ll have calculus I final exam next week. Thank you for the video!
Thanks!
I love it !
Wow ! Nice work. Not really difficult, but you have to be very persistent and confident.
Very clever solution!
I love the way he says square root
I believe it would be easier to solve using t=1/x substitution and then doing McLauren series expansion at point t=0
Yo i checked in the calculator and the limit is true!!
Itamar Rosen thanks!!!!!!
Damn, do you have a TI-89?
@@badhbhchadh most likely he didn't use a ti89, i typed it in like 2 minutes ago and it's still calculating LMAO
wolfram alpha gives it in 2 seconds tho
It means that you don't trust him.
Do you make these videos in Berkeley?
That was a very clever solution.
is a Indeterminaception XD
Eres un crack!!!.... cuando subirás ecuaciones en diferencia ???? (difference equations)
nice video man.really entartaining
This gave me the idea of an inverse limit. I can't really come up with a way for it to be used consistently. "Lim^-1 as x -> oo of 2" would be the notation. Maybe, instead of a constant, it would help do a function inside a function?
Do you mean "find a function that approaches this limit as x->inf"?
@@PixelSergey I'm not really sure.
I had that idea too, but then it quickly dawned upon me that the idea of an inverse limit is impossible. This is why:-
1) lim (2x/x)
x->oo
2) lim (x²/x)
x-> 2
3) lim (x)
x->2
All lead to the same result, but there's no way that we can list all possible limits which lead to 2. You will say, we cannot list all the answers of ln(-1), but that's different. There's at least a system by which we can do this. We only have to keep changing the number of rotations. This case is exactly the result of 'there are different types of infinities'. You can say that ln(-1) has countably infinite answers, while the idea of inverse limit has uncountably many answers.
@@createyourownfuture3840 is it different to log1(x)? I do admit the inverse is realistically useless. My guess is if it DID have a use, it would more likely for organization (ie. Making sure that you have to raise your "answer" to e in order to fully answer the question. Maybe something with catagorization theory (I heard it's a thing, and I have no clue what it's about minus the obvious)?
At 5:30, when using the x^2 pieces of sqrt to cancel the 2x, how can you ignore the remaining x^1 term, that would give some factor of sqrt(inf) in the denominator, which would send the limit to zero?
when solving limits where x-->inf , especially with functions that are one polynomial over another, a useful trick is to multiply both the numerator and the denominator by (1/x). keep doing this until the highest degree term is reduced to only a coefficient. then, when the limit is taken, those other terms end up being some number devided by infinity, which makes them zero. this is why he ignores all but the highest degree term; the other terms are reduced to zero.
That was a trip. Bravo
thats just awesome
Awesome!
Hi. You're so nice at math, especially on calculus. Can you do another video about algorithm? I'd like to see if you can do.
Wahyu Adi algorithm?
Whats the limit of this expresion when x goes to 0? It is a 0^0 situation
omg.....
Omg!! I've done it.
The answer is 1.
I just aplied LH two times and i ended with ln(L)=0 so L is equal to 1. I take a look with geogebra and im right! I can't believe this is the first limit i've solved
OH WOW! that's impressive!!!
Could you take the limit when x goes to negative infinty and work with complex numbers? I think that would be pretty dificult
(Sorry for the bad english)
@@gregorio8827 I, using my s___ty math ended up with 0, read my comment to see how I got it
How did you get rid of 2x under the radical
You should do a follow-up video doing this the right way, with the second-order Taylor approximation of the square root function.
First order gives you 1^infinity. Second order gives you (1-1/(2x))^x
I don't mind answering this question in my final exam but only if you are my calculus teacher, for I would then surely know that my efforts would be truly appreciated. Of course no other calculus teacher would even dream of such a vile limit, let alone put it on a final exam paper. Great explanation, great problem and as usual - that's it.
i have a rule made its called de' or eo rule (de)means take the derivative of all functions inside the parentheses and the derivative of the power as well, eo means take the integral ( the second step might not work, but you should try ot if the de step fails )
Brilliant..
The limit of inside without any calculations should be 1, you can use Murphy's Law for it.
Hahahahah
At the step where we get 2x/2x, can we not just conclude that the bottom part is strictly larger than 2x (since we ignored some parts and the square rote is strictly monotone increasing) and thus conclude that we are approaching 1 from the bottom, and since we are strictly smaller than 1, the limit of that to X is 0?
how do you know only to take x^2 terms from the denominator at 5:45?
i thought he was applying L'hopital's rule but i checked, and you cant solve it by that because the differentiation keeps getting more and more complicated
Thank god I finished all calculus courses before blackandredpen invented this monster
Rationalizing denominator for the final Answer?
ZAID SALAMEH why?
How?
scrt(e)/e still has an irrational denominator
ZAID SALAMEH "we're adults now"
jeromesnail (sqrt(e))^(-1)
What a brilliant sum ❤
This thing looks like the raidboss of calculus, but once we know its secret, it becomes easy(er)...... it still looks terrifying :D
I worked it out with the simple approximation (a+b)^.5 is approximately a^.5+b/(2a^.5) when a>>b.
Hey black red pen do you practice the problems before coming on the video.If no then 🤯 u blow my mind by such clean mathematical operations with out much mistakes,🤟
My calc 2 teacher, I already graduated why am watching these (as in I have seen over a hundred) I guess masters here I come lol.
You are great!
Somehow this video makes me feel like I just bought a used car from a guy who talks very fast.
i have no idea why i'm watching this, but it's so interesting i stayed the entire 22 minutes
wouldnt the 2x just make the left a lot bigger than the right side? So the whole term goes to infinity?
a very interesting limit indeed!
I think I have a challenge for you...
It's a question that was on the test for calculus monitor of my college, here in Brazil. It goes like this:
-Calculate: limit when n -> Infinite of ( 1/((√n).(√(n+1))) + 1/((√n).(√(n+2))) + 1/((√n).(√(n+3))) + ... + 1/((√n).(√(n+n))) ).
It might be a challenge, or not! It would be really nice to see you solving it in video, if possible!
Thanks for all your videos, they are very funny and inspiring!
As someone who is not very familiar with calculus, I feel like bprp is making up rules as he goes to make it look like he knows what he's doing (which he is)
Watched the video twice (and know calc) everything he does is correct
I just would have done it differently
omg dude you are brUral with your algebra
At 6:20 roughly I get confused; shouldn’t the limit be to 0 since we know both sqrt(x^2 +2x + 3) and sqrt(x^2 + 3) are both >x so the denominator is > 2x and the fraction is
Nicholas 4321 they are equivalent for large x values. more formally 2x/(sqrt(x^2+2x+3) +sqrt(x^2+3) = 2x/(x(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)))= 2/(sqrt(1+2/x + 3/x^2) + sqrt(1+3/x^2)) as x goes to infinity 2/x and 3/x^2 go to 0 (if this seems non-obvious, think about what happens when you divide something by a really big number, or look at their graphs) so we get 2/(sqrt(1)+sqrt(1))=2/2= 1
Nicholas Jenkins gotcha eliminating x from either the numerator or denominator seems to make it a lot easier thanks 😊
4:58 Which other video is he talking about?
Not quite understand of the video of time 5:10 so which video can explain this?Also not quite understand why no need absolute value?
I wonder if you can convert the indeterminate forms to a Magnitude and Angle
hi. I have a philosofical dilema with fractions and limits. When we have fractions of the form 1/x, and we want to calculate the limit (1/x)^y, when y goes to infinity, we always have (1^y)/(x^y), so we always have an indeterminate in the numerator. How have we to consider to solve this? Because i saw a lot when testig if a serie is convergent, in special geometric series, than is obviated this 1^inf. If you could bring me a spot light you will make me very happy. Thank you.
Not always, 1 is sometimes exactly 1 and you don't have to worry about the limit
lim x->infinity (1+1/x)^x->e; after we have 1^(infinity)=e
Amazing.
why do you rarely use series expansions? it seems so complicated without it when it could take 5 minutes without writing much.
\sqrt{x^2+2x+3}=x\sqrt{1+2/x+3/x^2}
=x(1+1/x+3/(2x^2)-1/8(2^2/x^2) + O(1/x^3))
=x+1+3/(2x)-1/(2x)+O(1/x^2)
\sqrt{x^2+3}=x\sqrt{1+3/x^2}
=x(1+3/(2x^2)+O(1/x^4))
=x+3/(2x)+O(1/x^3)
so overall the difference of square roots is 1-1/(2x)+O(1/x^2)
write it in exponential form and you get xln(1-1/(2x)+O(1/x^2))=x(-1/(2x)+O(1/x^2))=-1/2+O(1/x)
and thus your solution, exp(-1/2)
screenshot of a pdf version : i.imgur.com/uuHRur9.png
at 13:08 where does the x come from? didn’t you cancel out the two to get one in the numerator before? shouldn’t you get -sqrt(x^2+2x+3) rather than that with an x in front?
He only canceled the 2s on top and bottom of both fractions. The x is still on the top.
Wow, here it's even blackpenredpenbluepen.
Hi, you can solve this differential ecuation in terms of tanh? I solved in terms of ln but I'm interesing in the another result.
(dv/dt)=g-cv^2, v= velocity, g=gravity and c=k/m, k is a constant and m is the mass, the initial condition is V(0)=V0 (V0 is the velocity initial). Thank you :)
This is an example of the Abuse of Lopital's Rule not the use of it. Watch This Video fo learn More About the abuse of Lopital's Rule : czcams.com/video/3vAvpeqJkEs/video.html
After my math exam today, I'm definitely done with Mathe for the next 3 days xD