6 years and 2 weeks late, but I think i know a rough answer; The function ' e^(x * ln(x)) ' is continuous when x >=0. Additionally, the only part of the expression that is affected by the limit is the exponent, therefore we can bring the limit up. Another example would be in the expression lim (1 / (1+n)^1/n) n --> 0 The function is continuous if x>= -1 (I checked both these expressions on Desmos, you can as well) And the only part affected is the denominator. If you rearrange this with the limit on the denominator, (I'm not going to make you read that, just write it out or imagine it), you get something interesting; a limit definition of *e* . Simplify, and you get the limit approaches '1/e' as 'x' approaches 0
this is probably right, but someone correct me if I am wrong f being continuous means |x-a|< delta \; |f(x)-L| < epsilon => lim{x -> a} f(x) = L f( lim{x -> a} x ) = f(a) = L => lim{x -> a} f(x) = f( lim{x -> a} x ) input any generic continuous function instead of x and it still works (x is in fact a continuous function for example, that's why lim(x->a) of x = a in the third expression)
Lovely. You should make a video to summarize all main the basic tricks (transformation) ppl can do to solve limits (i.e. X = 1/1/X or A=e^ln(A) and son on)
We solved this exact problem when I was in high school. It was one of the coolest such problems. In fact, that inspired me to make the hypothesis that if the functions f(x) and g(x) are both smooth around a value of A, and f(A)=0 and g(A)=0, then the limit when x->A of f(x)^g(x) is 1.
@@mateidumitrescu238 It is actually true for analytic functions. There's a paper with a proof somewhere, but for the life of me I can't find it anymore (this is something really hard to google).
@@randoshmuckarias1296 If I had a link to the paper I would give it, but alas, I have lost it (and this is something that's pretty much impossible to google).
Luis G. I don't remember it all, but the general gist is that local linear approximation gets better the closest it is the the original point. And at 0 distance it will perfectly predict the value
I'm not really sure how you would go about proving this.... It would be sufficient to show that n^2 sin^2(n) > 1 for all n above some N. There may be some property of pi which makes this obvious but I don't know of it.
Can you do a follow up video showing why the low-point of this of this function is (1/e, e^(-1/e))?? Also on why there are two solutions for this function between x= 0 < 1/e < 1
blackpenredpen so if you would find time to make it, everybody will appreciate as many of us watch each of your videos as a movie with an interesting plot (btw im 17, Russian grade 11, know multivariable calculus, but have no idea why the hell I decided to learn all of this, maybe because of such amazing people like you and professor Leonard) sorry for mistakes 😃
What is the limit of this function as it approaches 0 from the left? Wolfram Alpha says that it is 1, but you've made me skeptical of that calculator. Btw, thanks for the video. You're videos never fail to entertain and to blow my mind. :)
It’s incorrect. As you approach 0 from the left, the limit DNE the same way if you approach form both sides. Note the domain of a function since x>0. Try this on your graphing calculator. If you look at some values that are less than 0, it may or may not see some dots and are not connected to each other. You could plot some negative integer values, but the problem is that if you put in some non-integer values, the answer may or may not have real values. That’s why we only care about positive bases similar to the way we graph for exponential functions. If the function is y=(abs(x))^x, the limit goes to 1 if we approach both sides as x goes to 0. Compare this with and without the absolute value.
I by no means know what I'm talking about, but I just don't worry about whether somethings complex-valued or not, I just let it be a number, and see what happens.
@@cameronbigley7483 I think the only limit equal to 1 and inf is the one when x tend to inf because the real part of the function is defined from 0 to inf. The limit when x tend to 0 is equal to 0 and when x tend to 1 is equal to inf.
It took me many hours to study and understand calculus. And what really helped me was the time that I actually tutored my classmates in calc. You can try the same to tutor/teach others and you wont believe how much more you will understand.
math is a huge branch of tree, unless you are a physics major/ engineering major, calculus is compulsory, if you are maths major, there’s a whole world out there without calculus, no need to force yourself into calculus. my tip for studying maths is do problems, and keep solving problems, then you will learn the pattern and techniques, just like, e^xlnx and know how to manipulate the equation to use Lhopital’s rule. no matter what subject you are taking, there is no running away from this advice
😕😕😕when you know your subject too well its ok to smile unlike others like me who are going through a phase of ups and downs of life cause exam is near and i am like😴😢😧😥😳😷😫no clue how to cover all the chapters?????
blackpenredpen I got 0, i don't know where my mistake is. First of all I said lim x->0+ x^x=L. Then I applied ln on both sides and got lim x->0+ xln(x)=ln(L) (after applying all logarythm properties) Then I split the limit into two using the product rule and got (lim x->0+ x)times (lim x->0+ ln(x))=ln(L). Then I elevated e to both sides of the ecuation and got e^(lim x->0+ x) times e^(lim x->0+ ln(x))=L. So the ln and e will cancel in the second factor and we'll get e^(lim x->0+ x) times (lim x->0+ x)=L. Finally we'll have e^0 times 0=L. 1 times 0=L, so L=0. :0
0^x is 0 for all x>0, x^0 is 1 unless x is 0. Depending on what the '0's are it is fairly simple to get 0^0 to approach either 0 or 1. In fact, depending on what the '0's are and how you approach the limit point, you can make 0^0 whatever you want.
is there a similar demonstration for limit x going to 0- ? I know imaginary numbers could get involved, but I think that would also go towards real 1 in the end...
hey! there is a quora question somebody asked about this formula, and somebody illustrated a 3D graph with the axes: X, Y(real), Y(imaginary), so it shows the magnificent shape of this graph when x
what if both x(the exponent x and the base x) have coefficients? should I apply the constant multiple law to the coefficient of base x then continue the process just like in the video?
![[#2024MAMA] G-DRAGON - 무제(Untitled, 2014)+POWER+HOME SWEET HOME+뱅뱅뱅+FANTASTIC BABY | Mnet 241123 방송](http://i.ytimg.com/vi/Ox29z5Nf1Uk/mqdefault.jpg)
Black Pen, Red Pen, Blue Pen, YAY!
Black pen, Red pen, Blue Pen, White Chalk
What if x tends to 0- pls any body explain
@@vsijahsjsjshsf599 it's a logrithmic function. It's isn't defined for x tends to 0-
@@shrijanbuchasia7366 thx
caneta azul azul caneta
2 videos the same day. Thank youuu:)
I just got out of the class where I learned this.
What class was that
@@wedeldylan University calculus LOL
Could you do a video on the whole thing about "we can bring the limit into the exponent because e^something is a continuous function"?
6 years late but I also wanna know about that... Can anyone help
6 years and 2 weeks late, but I think i know a rough answer;
The function
' e^(x * ln(x)) '
is continuous when x >=0.
Additionally, the only part of the expression that is affected by the limit is the exponent, therefore we can bring the limit up.
Another example would be in the expression
lim (1 / (1+n)^1/n)
n --> 0
The function is continuous if x>= -1 (I checked both these expressions on Desmos, you can as well)
And the only part affected is the denominator. If you rearrange this with the limit on the denominator, (I'm not going to make you read that, just write it out or imagine it), you get something interesting; a limit definition of *e* . Simplify, and you get the limit approaches '1/e' as 'x' approaches 0
this is probably right, but someone correct me if I am wrong
f being continuous means
|x-a|< delta \; |f(x)-L| < epsilon =>
lim{x -> a} f(x) = L
f( lim{x -> a} x ) = f(a) = L =>
lim{x -> a} f(x) = f( lim{x -> a} x )
input any generic continuous function instead of x and it still works
(x is in fact a continuous function for example, that's why lim(x->a) of x = a in the third expression)
You just saved my new assignments and taught me something new! (Didn't know about l'Hôpital's rule, researching that rn). Thank you!!!
Lovely. You should make a video to summarize all main the basic tricks (transformation) ppl can do to solve limits (i.e. X = 1/1/X or A=e^ln(A) and son on)
"Just to be legitimate..." You must note that x = e^(ln x) only when x>0, which in this case is true.
We solved this exact problem when I was in high school. It was one of the coolest such problems.
In fact, that inspired me to make the hypothesis that if the functions f(x) and g(x) are both smooth around a value of A, and f(A)=0 and g(A)=0, then the limit when x->A of f(x)^g(x) is 1.
@@mateidumitrescu238 It is actually true for analytic functions. There's a paper with a proof somewhere, but for the life of me I can't find it anymore (this is something really hard to google).
This is just a long way of saying limit of x^x when. X---> 0. Sounds like bullshit to me but maybe you can elaborate more
@@randoshmuckarias1296 sounds like you don't understand functions
@@randoshmuckarias1296 If I had a link to the paper I would give it, but alas, I have lost it (and this is something that's pretty much impossible to google).
You’re just like my professor but he retired, so I started watching you, man you’re great!
Looking sharp!
LOL
Thank you!!!!!!!!
Black Pen, Red Pen, Blue Pen, Green Pen, Yellow Pen, Pink Pen, Purple Pen, YAY!
How about limit of same function with plain 0?
I REALLY love your work ! I learn so many things. Thank you very much !!
Beautiful my man, very easy to understand proof.
Two classic tricks. limit of continuous function, and L'Hospitals. Neat.
Very impressive but you can show the proof of L'Hopital's rule?
Luis G. I don't remember it all, but the general gist is that local linear approximation gets better the closest it is the the original point. And at 0 distance it will perfectly predict the value
Proving de l'hopital in every case can be long. Haha
Luis G. It is too simple to show
It's simpler than most videos on this channel?
Can I just show a simple case?
Really... You are looking smart Sir.
I'm very impress with you because your teaching style is super. 👌👌👌👌👌👌👌👌👏👏👏👏👍👍👍👍
You are the best teacher I've already known.
lukas pereira. There are tons better teachers that I have met and they made me who I am today. Thank you for your nice comment tho!
I have a good challenge for you.
Test for convergence ∑(n=1 to inf) 1/(n^3 * (sin(n))^2).
Yan Nepomnyashi I imagine it is convergent
I'm not really sure how you would go about proving this.... It would be sufficient to show that n^2 sin^2(n) > 1 for all n above some N. There may be some property of pi which makes this obvious but I don't know of it.
Better you learn than judging whether other can solve something or not
this proof is absolutely brilliant.
Can you do a follow up video showing why the low-point of this of this function is (1/e, e^(-1/e))??
Also on why there are two solutions for this function between x= 0 < 1/e < 1
Please do a video on Cauchy's Residue theorem
th-cam.com/video/YWIseo5LwgQ/w-d-xo.html
Thank u so so much. YOU ARE VERY GOOD AT TEACHING A SUM
Cool example of limits
*stares at video
*stares at hands
...and I thought _I_ was good at math...
MORE INTEGRAL BATTLES PLEASE!!!!
I do have one idea in mind...
It's hella LONG........
blackpenredpen so if you would find time to make it, everybody will appreciate as many of us watch each of your videos as a movie with an interesting plot (btw im 17, Russian grade 11, know multivariable calculus, but have no idea why the hell I decided to learn all of this, maybe because of such amazing people like you and professor Leonard) sorry for mistakes 😃
5:53 subliminal advert for Coles Supermarkets?
Danana J I don't get it...
THANK YOU BOSS
great explanation, thanks a lot
What is the limit of this function as it approaches 0 from the left? Wolfram Alpha says that it is 1, but you've made me skeptical of that calculator. Btw, thanks for the video. You're videos never fail to entertain and to blow my mind. :)
It’s incorrect. As you approach 0 from the left, the limit DNE the same way if you approach form both sides. Note the domain of a function since x>0. Try this on your graphing calculator. If you look at some values that are less than 0, it may or may not see some dots and are not connected to each other. You could plot some negative integer values, but the problem is that if you put in some non-integer values, the answer may or may not have real values. That’s why we only care about positive bases similar to the way we graph for exponential functions. If the function is y=(abs(x))^x, the limit goes to 1 if we approach both sides as x goes to 0. Compare this with and without the absolute value.
Ohh it was Lhopital? My teachers just pronounced it as L-hospital bruh
Thank youu (from a 11th grader from Romania)
Very expressive but it is solve by l, hospital rule and let y equal this function and take log on both sides sir
Very creative!
loving your videos! please keep going :^)
I will!!
Similarly all x^(n×x) would give result 1 at x->0.
Nice. Most vids introducing limits don't give you this example.
Thanks for reminding me about L'Hopital's rule. I kicked myself a bit.
Love this video so much! Thanks!
Thank you so much. This Video was really helpful !!!
If x
it so easy to understand.i want little harder like (inf-inf)^(inf/inf).
Me parece brutal lo que haces..., muchas gracias
send x->0, let y=x^x, ln(y) = x*ln(x), then evaluate limit of x*ln(x). this approach seems a little bit easier
Another great video. Keep up the good work!
Thanks!!
U get me out from a rush
Please make videos on how to find area by integration
Please make a video on complex integration for beginners.
anuran chowdhury u mean complexifying the integral or use residue thm?
blackpenredpen Residue theorem. That will be great since complex analysis is presented in a very dry way.
Dr. Peyam did one video today (integral of 1/(x^4+1) from 0 to inf) !
I will have to find time to edit and upload soon.
I by no means know what I'm talking about, but I just don't worry about whether somethings complex-valued or not, I just let it be a number, and see what happens.
Hi, here a little challenge for u, can u demostrate the routh theorem of triangles
That is one spiffy suit!
Thanks!!!!!!
Lookin' suave mate!
Guy, you are magnificent!
your answer is most likely to be true if you answer either 0 or 1
Max Haibara not always
well I said most likely
Is there a way to calculate the limit without De l'Hopital's rule and without Taylor expansions?
Awesome job man!!
nicely done
Thanks!!!!!
U are so good man big fan 🙌🙌🙌
Can you help to solve this?
x^y = y^x
Where x and y are different.
Ex: 2^4 = 4^2
I'm pretty sure 2^4 and 4^2 are the only solutions to that.
x^y = y^x -> y*ln(x) = x*ln(y) -> ln(x)/x = ln(y)/y
Look at the function y(x) = ln(x)/x. For any 0
What do you mean by "limits of both ends"?.
Any solution x^y = y^x with x
There’s a video on this channel already answering this question
@@cameronbigley7483 I think the only limit equal to 1 and inf is the one when x tend to inf because the real part of the function is defined from 0 to inf. The limit when x tend to 0 is equal to 0 and when x tend to 1 is equal to inf.
Thank you.
JAM2020 question in physics
You are great mentor 😊
Wow that e is from the heart
6:04
you are smart, I' m a math major and calculus just so hard for me
It took me many hours to study and understand calculus. And what really helped me was the time that I actually tutored my classmates in calc. You can try the same to tutor/teach others and you wont believe how much more you will understand.
math is a huge branch of tree, unless you are a physics major/ engineering major, calculus is compulsory, if you are maths major, there’s a whole world out there without calculus, no need to force yourself into calculus. my tip for studying maths is do problems, and keep solving problems, then you will learn the pattern and techniques, just like, e^xlnx and know how to manipulate the equation to use Lhopital’s rule. no matter what subject you are taking, there is no running away from this advice
Hey you can also take logarithm on both sides and solve that will be easier imo
I feel weird inside, as if all this positive zero and infinity stuff is going against all my poor brain thought it knew...
Sir I'm your best fan... I love you sir
😕😕😕when you know your subject too well its ok to smile unlike others like me who are going through a phase of ups and downs of life cause exam is near and i am like😴😢😧😥😳😷😫no clue how to cover all the chapters?????
it can be solved using the expansion of taylor serie very easily
Nice suit 😁
How could you use a blue pen!? I loved this so much! Thank you
Amazing video
Hats off
Uhhh hahaha this was very easy, ir was in my calculus 1 final , btw , Nice suit!
I found this comment earlier, "Why did the square on x disappear?"
Yee, my intuition didn't fail me.
0^0 isn't always 1.
You can check out one of my recent vids for example
Brilliant
Black Pen Red Pen please do i factorial. Thanks!
We had to do this without L'Hospital, that was cruel.
Arteyyy how did u do it without the LH rule?
blackpenredpen I got 0, i don't know where my mistake is. First of all I said lim x->0+ x^x=L. Then I applied ln on both sides and got lim x->0+ xln(x)=ln(L) (after applying all logarythm properties) Then I split the limit into two using the product rule and got (lim x->0+ x)times (lim x->0+ ln(x))=ln(L). Then I elevated e to both sides of the ecuation and got e^(lim x->0+ x) times e^(lim x->0+ ln(x))=L. So the ln and e will cancel in the second factor and we'll get e^(lim x->0+ x) times (lim x->0+ x)=L. Finally we'll have e^0 times 0=L. 1 times 0=L, so L=0. :0
There’s no algebra method to this. l’Hopital’s way is the only method to work it out for one of the seven indeterminate forms.
Thanks
Can u giving more videos that talk about how to solve limits without use the L’hopital theorem ?teacher!!!
This is why i think 0^0 should equal 1
probe the property of continuous function yay!!! fun fun
Thanks sir g
Could there be another limit of the form 0^0 where the answer is not 1?
David Espinoza yes. Prob do a vid next week
x^0 is always 1
HS Antonin but 0^x is always 0.
Nope, 0^0=1
0^x is 0 for all x>0, x^0 is 1 unless x is 0. Depending on what the '0's are it is fairly simple to get 0^0 to approach either 0 or 1. In fact, depending on what the '0's are and how you approach the limit point, you can make 0^0 whatever you want.
Amazing
Hey, some very intriguing maths, I'm currently studying maths at college and was wondering do you have any good books on calculus, especially limits.
is there a similar demonstration for limit x going to 0- ? I know imaginary numbers could get involved, but I think that would also go towards real 1 in the end...
very very very very fun!!!!
What is the domain and range of the function (x^x) ?? 🤔
Cool dress
Can't it be only differentiating x^x = x.x^x-1 and putting the limit? Since it'll be 0 × x^x-1 = 0 for any case
not sure I understand how one can change to the lim of x^x in the exponent .
this is so effing cool
Can you give a feel? I can't able to understand it by graph
Love it
Whats happens when same question but x tends to 0^- ?
all you math nerds watching this have a nice day
Dennis Brinken
We will.
You watched the video too. So have a nice day to you as well.
Please make a video of lim x^x^x^...... as x->0+ ?
I found him today, Can i hug him?
hey! there is a quora question somebody asked about this formula, and somebody illustrated a 3D graph with the axes: X, Y(real), Y(imaginary), so it shows the magnificent shape of this graph when x
Legitimate!
what if both x(the exponent x and the base x) have coefficients? should I apply the constant multiple law to the coefficient of base x then continue the process just like in the video?
He could have just done e^0 * e^-inf which is just 1-0 which is just 1