Great video. Informative and enthusiastic. I like math, but people who don't like math might start liking it if they saw how much fun it looks like you're having in these videos.
I have watched a couple of videos of yours and I must say your enthusiasm is contagious. All this while keeping things straightforward, neither oversimplifying things not complicating them. It is good to revisit the fundamentals so watching these videos from time to time makes sure I don't lose grip on the most basic techniques in maths. Love from India. Keep doing what you're doing 😊
I wasn't finding any brazilian video that could help me with this problem in a simple way, but u came to save me. Despite being a video with another language, it was an amazing explanation !!!! Thanks for ur help😊
A possibly more precise way to say it: the limit of f(g(x)) as x-->a (for some scalar a in [-inf,inf]) equals f evaluated at the limit of g(x) as x-->a if * the limit of g(x) as x-->a exists, equalling some value b, AND * f(x) is continuous at b [or in the case that b equals +- inf we require that the limit of f at +- inf exists] In any case: a very interesting limit in this video and a nicely explained solution! Thanks for your efforts!
It's a crime that TH-cam doesn't recommend you more. Found out about you about 2 days ago and, gahdamn, like everyone in your comments say, you're *awesome*. I really like the little pauses you do at times, it helps (at least) me digest whatever you just did better.
Assign x := 1/t, so that: f(x) = f(1/t) Then find instead, Lim: t --> 0 | f(1/t) (Which approaches the same limit value) You get:. Lim: t --> 0; (1/t)^t = 1/t^t. Since Lim: t --> 0; t^t = 1, then 1/t^t --> 1/1 = 1. Therefore: Lim: x --> infinity | f(x) = 1.
I think we did this in calc using a sequence, and showing that n^(1/n) has same limit as n/n+1 and then in continous case we used some theorem. Never thought of this done this way. cool stuff.
oh hey, this exact method proves the limit of (1+1/x)^x as x goes to infinity is e, nice (i know it's also often given as the definition of e, but if we define e by its calculus properties instead, this proves they're the same value, which is nice)
We try to avoid 'inserting' infinity into anything. Always use infinity as limits. (Infinty)^0 is indeterminate and therefore can only be computed as a limit.
can be rewritten as x to the power of 1/x, since x is an increasing very big number, the result of 1/x will be closer and closer to zero. Anything to the power of 0 is 1, so the entire expression will evaluate closer and closer to 1
Excellent video! One thing that you said is not always true! Consider the limit as x-oo of x/(2x). Clearly both the numerator and denominator both go to infinity AND the denominator is larger than the numerator. However, the limit goes to 1/2 not 0 as you basically said it would (Look at 7:10 - 7:30).
Hello sir, i want to ask something about limits which approach to infinity. When we're trying to solve a infinity limit problem we're trying to avoid inserting infinity because it's not exactly a number as we all know. So does that mean key of solving infinitiy limits is actually trying to only keep "1/x" expressions in the equation of function? So basically is only thing we can say not indeterminate "1/x" (Except another things like infinity+infinity=infinity or infinity*infinity=infinity etc.)?
Professor, can we argue that at step 2, "the limit of x to the power of 1/x as x goes to infinite" the limit is 1? for the limit of 1/x is 0 and any number x (including it being infinity) raised to the the power of 0 is 1?
So, the question is: is x^0 equal to 1? I don't agree with that. The definition of a^b is "take 0 and a multiplied with itself for b times". If you place b=0 what do you get? 0 or 1?
I don't get one move here. Why is it justified to move from the ratio of the two ln functions here to the ratio of their derivatives? How do we know that move is legitimate? Perhaps I'm merely slow, but it is not obvious to me that one can do that.
1th root of 1 is 1, √2 = 1.4142, oh it's increasing. Mmh. ³√3 = 1.4. Oh so it decreases between 2 and 3. I have NO clue, but I'm gonna bet my family the maximum value is the eth root of e 😂
If this is constructive criticism, it is more effective if you highlight areas that need improvement and be specific. Also, realize that everyone has at least one flaw. So, be gracious in criticizing others, especially if you are not their coach. I have also learned to suggest things to others while recognizing that my suggestion is optional. Hope you read this and really tell me the meaning of blagging so I can learn. We Never Stop Learning!
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Great video. Informative and enthusiastic. I like math, but people who don't like math might start liking it if they saw how much fun it looks like you're having in these videos.
Best Maths tutor on the net.
I have watched a couple of videos of yours and I must say your enthusiasm is contagious. All this while keeping things straightforward, neither oversimplifying things not complicating them.
It is good to revisit the fundamentals so watching these videos from time to time makes sure I don't lose grip on the most basic techniques in maths.
Love from India. Keep doing what you're doing 😊
Thanks for watching!
Couldn't be more clear and articulate!
Gem! Without these playlists life could've never been this easy.
I wasn't finding any brazilian video that could help me with this problem in a simple way, but u came to save me. Despite being a video with another language, it was an amazing explanation !!!! Thanks for ur help😊
To avoid misunderstandings, instead of "x^1/x," it should be written as "x^(1/x)". So, the exponent 1/x should be enclosed in parentheses as (1/x).
Verdade, não achei nada em português, esse cara é muito bom
great video, I tried watching others but didn't understand shit, you put things into perspectives, thank you man
👍👍continuity! I love the sound of chalk on a blackboard, a sound that will be soon lost to history.
Love the energy of this dude
the limit of a function is the function of the limit as long the function is continuous.
It would be important to specify where these functions are expected to be continuous, as they do not need to be continuous everywhere.
Which function, both of them?
A possibly more precise way to say it: the limit of f(g(x)) as x-->a (for some scalar a in [-inf,inf]) equals f evaluated at the limit of g(x) as x-->a if
* the limit of g(x) as x-->a exists, equalling some value b, AND
* f(x) is continuous at b [or in the case that b equals +- inf we require that the limit of f at +- inf exists]
In any case: a very interesting limit in this video and a nicely explained solution! Thanks for your efforts!
That's the sequential definition of continuity.
Fascinating story! Many thanks for your video! 😃
Thank you.
Nice problem and excellent explanation. Well done. Thanks and keep up the good work.
best teacher ever
Love your videos, you are really good at explaining!
It's a crime that TH-cam doesn't recommend you more.
Found out about you about 2 days ago and, gahdamn, like everyone in your comments say, you're *awesome*.
I really like the little pauses you do at times, it helps (at least) me digest whatever you just did better.
Love this channel!
This was super helpful, thanks! You have a really nice attitude and a great explanation. Thanks again:)
I'm glad you find them helpful.
"Never stop learning, because if you stop learning, you stop living" ="the limit of a function is the function of the limit"
What a great way to present mathematics
Super film. Rozwiązanie jest genialne. Nigdy bym na to nie wpadł.
Assign x := 1/t, so that: f(x) = f(1/t)
Then find instead, Lim: t --> 0 | f(1/t)
(Which approaches the same limit value)
You get:. Lim: t --> 0; (1/t)^t = 1/t^t.
Since Lim: t --> 0; t^t = 1, then 1/t^t --> 1/1 = 1.
Therefore: Lim: x --> infinity | f(x) = 1.
that really helped me to understand the problem. Thank you sir! I appreciate your efforts!!
I love your delivery.
Simply great!!!
Nice! e saves the day! 😊 Thanks for a great video.
Thank you, Doctor, i an going to soak up his teachings. It is so illuminating and illustrating the beauty of mathematics. Happy new year, doctor.
This is great my man
The light shade on the blackboard sometimes makes it difficult to see the board . Please change the settings.
Very good. Thanks 🙏
Wonderful sir
Those who stop learning stop living
So
Never stop learning
Great saying
You did not say "Now on to the video" before you did your musical interlude. I missed it.
Oh the explanation was completely awesome !!!
I think we did this in calc using a sequence, and showing that n^(1/n) has same limit as n/n+1 and then in continous case we used some theorem. Never thought of this done this way. cool stuff.
Thank you very much.
Well done!
Great fun AND instructive
Thank you so much
Muito bom, parabens
Bella spiegazione, grazie
oh hey, this exact method proves the limit of (1+1/x)^x as x goes to infinity is e, nice
(i know it's also often given as the definition of e, but if we define e by its calculus properties instead, this proves they're the same value, which is nice)
Why not just insert 'infinity' for 'x' in '1÷x' in the fractional exponent, resulting in 'x^0', thereby reaching the same result?
We try to avoid 'inserting' infinity into anything. Always use infinity as limits. (Infinty)^0 is indeterminate and therefore can only be computed as a limit.
No, it cannot be done. Infinity is not a number, it's only a concept.
I Had the Same Idea. With your explanations, I have a question. Isn't there a Proof that anything to the power of 0 is 1?
Can you make a video of the actual application of limits in some real world physics problems?
Nice solution!
can be rewritten as x to the power of 1/x, since x is an increasing very big number, the result of 1/x will be closer and closer to zero. Anything to the power of 0 is 1, so the entire expression will evaluate closer and closer to 1
Amazing
Excellent video! One thing that you said is not always true! Consider the limit as x-oo of x/(2x). Clearly both the numerator and denominator both go to infinity AND the denominator is larger than the numerator. However, the limit goes to 1/2 not 0 as you basically said it would (Look at 7:10 - 7:30).
i was searching for this comment
So good
very nice
Hello sir, i want to ask something about limits which approach to infinity. When we're trying to solve a infinity limit problem we're trying to avoid inserting infinity because it's not exactly a number as we all know. So does that mean key of solving infinitiy limits is actually trying to only keep "1/x" expressions in the equation of function? So basically is only thing we can say not indeterminate "1/x" (Except another things like infinity+infinity=infinity or infinity*infinity=infinity etc.)?
Yes
To avoid misunderstandings, instead of "x^1/x," it should be written as "x^(1/x)". So, the exponent 1/x should be enclosed in parentheses as (1/x).
Thank you. But x at power of x=x. Hiwvyou solve, sir?
I watched solution with squeeze theorem
Wouldn't it have been much easier to stop at the second step "X raised to (1/X)" and say that it would be just like raising any number by 0? Hence 1?
X raised 1/X, but when X tends to infinity, it's like infinity to the power zero, so it's undefined.
But the limit gives ∞^(1/∞) since 1/∞ is 0 then it’s ∞^0 which is 1. I know infinity and zero don’t go well together but that works right?
Professor, can we argue that at step 2, "the limit of x to the power of 1/x as x goes to infinite" the limit is 1? for the limit of 1/x is 0 and any number x (including it being infinity) raised to the the power of 0 is 1?
So, the question is: is x^0 equal to 1? I don't agree with that. The definition of a^b is "take 0 and a multiplied with itself for b times". If you place b=0 what do you get? 0 or 1?
thanks!
e does not work as a constant here, and you cannot always move limits around functions
lim x -> infinity x^1/x = x^lim x -> infinity 1/x = x^0 = 1, I think this is more simple?
This problem has been discussed more than the weather. Is 0^0 0 or 1? My vote is 1.
nice comment!
I don't get one move here. Why is it justified to move from the ratio of the two ln functions here to the ratio of their derivatives? How do we know that move is legitimate? Perhaps I'm merely slow, but it is not obvious to me that one can do that.
It is the l'hopitals rule, check it out on the internet there is a proof of that statement.
👍👍
Couldn't follow how the (e) was exponentiated to the limit...
That is a limit law. As long as a function is continuous, 'the limit of the function is the function of the limit'
Absolutely 💯
What Newton said
👍👍👍👍👍👍👍👍👍👍👍👍👍👍
Maths could be damn entertaining with Newton.
L=1
🤩
couldn`t understand how you differentiated (ln x)
Derivative of x divided by x
No, It won't that It can go below 1?!
limxinf x^(1/x)=inf^0=1
Don't we have to prove "logX/X goes to 0" ? Indeed it must be 0 but...
The natural number e is omnipresent 😂
1th root of 1 is 1, √2 = 1.4142, oh it's increasing. Mmh. ³√3 = 1.4. Oh so it decreases between 2 and 3. I have NO clue, but I'm gonna bet my family the maximum value is the eth root of e 😂
I have watched a few of your videos and feel that you are blagging it sometimes
Blagging it? Wdym
If this is constructive criticism, it is more effective if you highlight areas that need improvement and be specific. Also, realize that everyone has at least one flaw. So, be gracious in criticizing others, especially if you are not their coach. I have also learned to suggest things to others while recognizing that my suggestion is optional. Hope you read this and really tell me the meaning of blagging so I can learn. We Never Stop Learning!
Thank you!