You explain these things in such a clear and concise way that me being a high school student, I still get some knowledge out of these videos. Hope I have some professors as good as this next year
Hi, sir! Long time fan here. I watch all your videos for fun and for study. Your calculus videos are awesome and very entertaining. Got very excited when you're doing the infinite series, one of my fav topics in calculus II. Can I make suggestions for future topics for your videos? Can you make videos about the ever interesting stuff about the Bernoulli numbers, gamma function, Zeta functions, q series, Ramanujan Summation? Any stuff related to Real, Complex and Analytical Number Theory with be great for me. Thanks and keep the genius coming!
Joshua Garcia hi josh, thank you so much for your nice comment. I may not be able to do them anytime soon tho since I have a lot of topics that I want to cover for my students first. But whenever I have time, I will definitely squeeze in some videos about random math problems for fun!
@@blackpenredpen I took the natural log but then didnt think of lhopitals rule..isnt there ankther way to do it without lhopitals? I really hope you can please respond.
I'm currently taking Precalculus in High School, but the way you explained it somehow makes me understand it! The videos you've been making are awesome, and I hope you continue showing us things as cool as this!
I thought I recognized the function. After you got the answer, I noticed this is what we called the "Pert" function for continuously compounding yearly interest. V = P(1 + R/X)^(T*X) P=initial principal R=interest rate T=number of years the money is in the bank X=the number of times a year you compound the interest (1 for yearly, 12 for monthly, 365 for daily, etc) Compounded continuously, X=infinity, so V = P*e^(R*T)
Yup! e is just the interest rate of continuous compounding over 1 unit of time (let's say year), assuming non-compounded 100% interest over 1yr. If you increase the yrs from 1 to b, you get e^b of course, since you just grow multiplicatively by e each yr. And if you increase the rate from 1 to a, you again get the same effect, because you've reduced the time it takes to get to 100% interest w/o compounding to 1/a yrs. Meaning 1 yr at rate a is identical to a yrs at rate 1, so e^a. Combining the two gets e^ab. (You can check that, by symmetry, the same effect happens when a
I love the informal pet names that teachers give to certain uhh facts. It's fun and it actually helps students remember important concepts! For instance, my calculus teacher called the chain rule "THE MAGIC". She always said, REMEMBER THE MAGIC.
I think I have a simpler way of doing this.... Since (1+x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3!........ putting x= a/x and n=bx we get 1 + ba/1! + (ba)^2/2! - b(a)^2/x+ ... and now all the terms with x in the denominator will become 0 as x->infinity so we are left with 1 + ab/1! + (ab)^2/2! + (ab)^3/3!.... this is just like the wxpression for e^x e^x = 1 + x/1! +x^2/2!....... therefore the answer is (e)^ab
@@leif1075 you find it with developemtn in serie : exp verifies the differential equation exp = d/dx (exp) so you search a solution serie of general term (a_n x^n) and you find exp(X) = Σ x^n/n! So e = Σ 1/n!
or we can say that (1+a/x)^bx = e^(bx*ln(1+a/x)) so when x approaches infinity a/x approaches zero and we can use ln(1+x) = x+ o(x) so L = e^(b*x*a/x) = e^ab !
We can use lim x tends to infinity (1+a/x)^bx as e^lim x tends to infinity a/x divide by bx So e^ limit x tends to zero a/x*bx Then a /x multiple by bx and x will be cancel our And it becomes e^ab
Can't you do it much simpler if you know what e is? what you know: lim x->∞ (1+a/x)^x = e^a and g^pq = (g^p)^q so... lim x->∞ (1+a/x)^bx = lim x->∞ ((1+a/x)^x)^b = (e^a)^b = e^ab
I'm just wondering: who would encounter this problem before learning what e is I mean, you use Ln, and limits, even the property that Ln is continuous, and l'hopital's rule, so they must have some idea.
... The " Special " Fact (a = 1 and b = 1) --> lim(x-->inf)(1 + 1/x)^x = e lim(x-->0)(1 + x)^1/x = e ... (lol) ... thank you for a great and very clearly understood presentation, Jan-W
The answer if obvious if you recognize this as the limit of the compound interest formula as the compounding period goes to infinity. P(1+r/n)^(t*n) - > Pe^(rt)
I enjoy your videos but I'm slightly worried by this one as it could mislead people. You haven't shown the limit is e^(ab). Rather, you have shown that if there is a limit then it must be e^(ab). I would have liked you to have mentioned why the limit must exist.
Yes, that's very fine. Just keep posting em and me and my mates'll watch them. We're very entertained. Thanks so much for these videos. You're way fun and captivating than my professors!
Man, I really wish I'd seen your videos during A levels. You are so fun in your explenations. Now I'm studying mathematics at uni and while your vids are great I don't get the same satisfaction due to the fact that I can tell the answers myself quite easily. :(
Other way to do it is to simply evaluating the limit of e^(bx * ln(1 + a/x)). Doing L'Hôpital (of course doing bx = 1/(1/bx), else it would not make sense) would be essentially the same as you did, just without equalling to anything to get an answer; OR you can use the transformation lim(x->c) ln(f(x)) = lim(x->c) (f(x) - 1) if and only if lim(x->c) f(x) = 1: e^(lim(x->∞) bx (1 - a/x - 1)) = = e^(lim(x->∞) bx * a/x) = = e^(lim(x->∞) ab) = e^ab
The result still holds if we replace a & b with sequences that converge to a & b, respectively. This is often encountered in statistics. See the following video. ruclips.net/video/k1BhXt9DgA4/видео.html
Why don't you utilise the definition of e.... lim x->inf (1+1/x)^x =e lim x->inf (1+a/x)^bx =lim x->inf [1+1/(x/a)]^[(x/a)(ba)] ={lim x->inf [1+1/(x/a)]^(x/a)}^ba =e^ab which is much easier....
The reason is I can show this video to my students if they have to do something like, lim as x goes to inf, (1+1/x^2)^x This limit is 1. www.wolframalpha.com/input/?i=limit+as+x+goes+to+inf,+(1%2B1%2Fx%5E2)%5Ex
Can anyone correct me if I'm wrong, but this formula should still work even if the denominator x, has a constant being added or subtracted, for example x + 2 or x - 5, cause at the end, it will eventually be zero and you will be left with (ab)/(1 + 0), which is just ab
Okay, so you used a natural logarithm to get bx out front. What if you used any other logarithm (such as log base 10) to do so? Would that not change the answer? Are you only allowed to make everything in your equation the exponent of the base of the logarithm you want to cancel out if it is a natural logarithm?
I would not put the limit L because we do not know if the limit exists a priori..maybe this techniques gives a correct result in this case ,but in general it is not correct(for pathogenic functions)..If f(x) is the function mentioned in the video i would write: f(x)=e^ln(f(x)) and the compute the limit of ln(f(x))
how about if it is a number other than 1 e.g lim x-> 0 ( 2 + a/x) inside the limit (i presume the answer doesn’t involve e any more? maybe with a natural log of an integer?)
I feel like doing the derivative of the numerator and demonstrator doesn't guarantee continuity unless the function in the numerator and denominator are the same. otherwise the delta of one will grow at a different rate and the fraction will no longer represent the same ratio. Same reason why 1/1 = 1^4/1^4 but 4/1 =/= 4^3/1^3 you can't just apply operations willy nilly on numerator and denominator.
I call this 'The Magical Limit Escape' When all the checks, La Hospital Rule or any other rule don't lead to me any conclusion, this formula (according to you) works as a magic and leads to the exit way!!!
Thanks for the great explanations!!! BTW can you also make videos for improper integrals for finding the values of p and q in the condition of convergence?
Or you could use the well known limit of ln(1+x)/x when x tends to 0. If you don't know that this is 1, you can see this this as the derivative of x\mapsto ln(1+x) in 0. Way easier than the 'hospital rule' that is something students absolutely don't understand well enough to use correctly.
But don't understand most probably. To the majority of student, it is like 'magic'. For proof, often they use is even if f(0) eq 0. In any case, I think my suggestion is a bit more elementary, like you wished (do itt from scratch). Especially, using the 'hospital rule' with the indentity in the denominator seems a bit complicated for nothing :p
Would there be a similar limit for complex z, i.e. for some sequence z_n such that for all n(in Natural numbers), there exists some N such that for all m>N, |z_m|>n, the limit lim(n->∞)(1+a/z_n)^bz_n exists and is equal to e^ab?
You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.
can you clear my doubt, that 1 to the power anything is 1 because if you keep multiplying 1 to itself its not gonna change. so why 1 to the power infinity is not one?
As I mentioned above: You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.
I still don't understand why ln(lim(f(x))) = lim(ln(f(x))). You said that ln was a continuous function, which I understand, but I don't understand how that yields the swap correct algebraically.
no as the limits result would change based on how you evaluate it ? this doesnt make sense. this is because e is the only number you can differentiate in an exponential function so you have to take take ln not log 10.
@@JensenPlaysMC The function f(x)=logx is an "1-1" function. This means that x1=x2 f (x1)=f(x2). So when logx1=logx2 then and only then x1=x2. We both agree on that. In this particular situation x1=L and x2 is our limit. Why this doesn't apply here? Both logx and lnx are functions that work the same way. The difference is the base. But I don't get why e works and 10 doesn't. I'm a student, last year before university, maybe my knowledge isn't enough, but I would like to know that. Is there any link that explains it? Thanks in advance.
@@ΒαγγέληςΚωστούλας-τ2ν because when answering this question you need to use l'hospitals rule as you have an inteterminate form( differentiate numerator and denominator) in order to make sense of the limit. now. let me ask you this. when you do log 10, ( like he did to ln,e). you need you differentiate it. how do we differentiate log 10? the reason we use ln is because to differentiate any exponential function you need to use the property that e is the derivative of it self, in order to determine that the derivative of ln (1+a/x) is 1/(1+a/x) , and then use chain rule. now. you could use log 10, but then you would have to find the derivative of that which is hard to do. as if you use first princibles you see that any number differentiated apart from e^x has another string of numbers times with it
and you can express this is terms of e, much like the derivative of 2^x is (2^x * ln 2). so using any base other than e is not really good as you can express any other base in terms of e
@@JensenPlaysMC Thanks again for your reply. I think I sorted it out now. Is this a consequence of the fact that the derivative of a function is unique?
I envy your students. You are an amazing teacher.
Aaron Hollander thank you!
I really love your enthusiasm. Thank you for your videos and the magnificent lessons!
Marian P. Gajda thank you for your nice comment.
You explain these things in such a clear and concise way that me being a high school student, I still get some knowledge out of these videos. Hope I have some professors as good as this next year
Thank you. I hope you the best too. Are you in 11th or 12th grade?
blackpenredpen 12th grade, currently learning about optimization in calculus. It's my favourite subject:)
Uchiha Madara I see! That's great!! I like optimizations too and should definitely make videos on then soon!
Hi, sir! Long time fan here. I watch all your videos for fun and for study. Your calculus videos are awesome and very entertaining. Got very excited when you're doing the infinite series, one of my fav topics in calculus II. Can I make suggestions for future topics for your videos? Can you make videos about the ever interesting stuff about the Bernoulli numbers, gamma function, Zeta functions, q series, Ramanujan Summation? Any stuff related to Real, Complex and Analytical Number Theory with be great for me. Thanks and keep the genius coming!
Joshua Garcia hi josh, thank you so much for your nice comment. I may not be able to do them anytime soon tho since I have a lot of topics that I want to cover for my students first. But whenever I have time, I will definitely squeeze in some videos about random math problems for fun!
@@blackpenredpen I took the natural log but then didnt think of lhopitals rule..isnt there ankther way to do it without lhopitals? I really hope you can please respond.
I'm currently taking Precalculus in High School, but the way you explained it somehow makes me understand it! The videos you've been making are awesome, and I hope you continue showing us things as cool as this!
I thought I recognized the function. After you got the answer, I noticed this is what we called the "Pert" function for continuously compounding yearly interest.
V = P(1 + R/X)^(T*X)
P=initial principal
R=interest rate
T=number of years the money is in the bank
X=the number of times a year you compound the interest (1 for yearly, 12 for monthly, 365 for daily, etc)
Compounded continuously, X=infinity, so V = P*e^(R*T)
Yup! e is just the interest rate of continuous compounding over 1 unit of time (let's say year), assuming non-compounded 100% interest over 1yr. If you increase the yrs from 1 to b, you get e^b of course, since you just grow multiplicatively by e each yr. And if you increase the rate from 1 to a, you again get the same effect, because you've reduced the time it takes to get to 100% interest w/o compounding to 1/a yrs. Meaning 1 yr at rate a is identical to a yrs at rate 1, so e^a. Combining the two gets e^ab. (You can check that, by symmetry, the same effect happens when a
I love the informal pet names that teachers give to certain uhh facts. It's fun and it actually helps students remember important concepts! For instance, my calculus teacher called the chain rule "THE MAGIC". She always said, REMEMBER THE MAGIC.
Yea, I find it super helpful too!!
Ah, that's why he uses "fish" for the W function...
I think I have a simpler way of doing this....
Since
(1+x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3!........
putting x= a/x and n=bx
we get
1 + ba/1! + (ba)^2/2! - b(a)^2/x+ ...
and now all the terms with x in the denominator will become 0 as
x->infinity
so we are left with
1 + ab/1! + (ab)^2/2! + (ab)^3/3!....
this is just like the wxpression for e^x
e^x = 1 + x/1! +x^2/2!.......
therefore the answer is
(e)^ab
Makes sense
Man, that's brilliant!
But what if you didn't know or haven't proved that infinite expression for e?
@@leif1075 you find it with developemtn in serie : exp verifies the differential equation exp = d/dx (exp) so you search a solution serie of general term (a_n x^n) and you find exp(X) = Σ x^n/n! So e = Σ 1/n!
it's known that lim(x-->inf) (1+t/x)^x = e^t
now if b>0 just replace by (1+(ab)/(xb)) and replace x by bx, you get e^(ab) I
if b
"The fact" is actually a very natural property :D
Here’s a faster way:
lim((1+a/x)^(bx))
= lim((1+a/x)^x)^b
= (lim((1+a/x)^x))^b
(1+a/x)^x = e^a when x appr infinity
= (e^a)^b
= e^(ab) //
love how you call it "The Fact"! 😂
Ahmed Anwari lol! Thank you. My students love it too! It works
or we can say that (1+a/x)^bx = e^(bx*ln(1+a/x)) so when x approaches infinity a/x approaches zero and we can use ln(1+x) = x+ o(x) so L = e^(b*x*a/x) = e^ab !
I really love your videos btw
Imaspammedboy thank you
I wish you had been my math teacher. Great work. I enjoy your videos even though I haven't had a math class in many decades.
"ln-ded both sides" what did you just say? 10:40
We can use lim x tends to infinity (1+a/x)^bx
as e^lim x tends to infinity a/x divide by bx
So e^ limit x tends to zero a/x*bx
Then a /x multiple by bx and x will be cancel our
And it becomes e^ab
we can actually do this without l'hopitals rule by the way your way of explaining is so good!
Can't you do it much simpler if you know what e is? what you know:
lim x->∞ (1+a/x)^x = e^a
and
g^pq = (g^p)^q
so...
lim x->∞ (1+a/x)^bx = lim x->∞ ((1+a/x)^x)^b = (e^a)^b = e^ab
mrBorkD true. But as I said "do it from scratch" without even knowing "your fact"
I'm just wondering: who would encounter this problem before learning what e is
I mean, you use Ln, and limits, even the property that Ln is continuous, and l'hopital's rule, so they must have some idea.
mrBorkD true. But I know my students. This approach is more suitable for most of them
OK. well I can't argue with you there
Exactly how I did this :D
I mean that way it can be proven with school mathematics (Binomial theorem and power rules, that's all) and done.
... The " Special " Fact (a = 1 and b = 1) --> lim(x-->inf)(1 + 1/x)^x = e lim(x-->0)(1 + x)^1/x = e ... (lol) ... thank you for a great and very clearly understood presentation, Jan-W
you and my cal 2 professor are making cal 2 very easy!! You guys are the best:)
The answer if obvious if you recognize this as the limit of the compound interest formula as the compounding period goes to infinity. P(1+r/n)^(t*n) - > Pe^(rt)
We can generalise the function of 1^infinity as
Lin X-->inf f(x)^g(x) = e^[{f(x)-1}g(x)]
You could also have evaluated that limit using a series expansion: ln(1+a/x) = a/x to first order, which gives you the answer in a couple more lines.
I think I need a channel teaching the first few billion years of maths...
Even if I was doing this right I'd feel like I was doing this wrong
Put = ay and the result follows at once. I guess he wanted to show the principle.
In my personal mnemonics i'm calling this the 'E-fact'
I enjoy your videos but I'm slightly worried by this one as it could mislead people. You haven't shown the limit is e^(ab). Rather, you have shown that if there is a limit then it must be e^(ab). I would have liked you to have mentioned why the limit must exist.
Yes, that's very fine. Just keep posting em and me and my mates'll watch them. We're very entertained. Thanks so much for these videos. You're way fun and captivating than my professors!
The cool accent, mic and apparent joy from calculus make this video. As all others from this guy. Goddamn do I subscribe!
LOL! Thanks!
Can't we also substitute the variables to get the( definition of e) ^ab
Man, I really wish I'd seen your videos during A levels. You are so fun in your explenations. Now I'm studying mathematics at uni and while your vids are great I don't get the same satisfaction due to the fact that I can tell the answers myself quite easily. :(
If you sub in a and b to be 1 then you get the equation used to find e (1+(1/n)^n which makes sense coz it’s a^(1•1) which is e.
This Fact, was indeed very well done
You never fail to impress me!
Thanks!
Hey, we can take the a in bottom and at exponent we do a/a. Like this we have direct e^ab
Setting a, = b = 1 is a nice way of getting to compound interest tending to e
yea
... that's THE FACT Jack ... !
WJL what jack?
Other way to do it is to simply evaluating the limit of e^(bx * ln(1 + a/x)). Doing L'Hôpital (of course doing bx = 1/(1/bx), else it would not make sense) would be essentially the same as you did, just without equalling to anything to get an answer; OR you can use the transformation lim(x->c) ln(f(x)) = lim(x->c) (f(x) - 1) if and only if lim(x->c) f(x) = 1:
e^(lim(x->∞) bx (1 - a/x - 1)) =
= e^(lim(x->∞) bx * a/x) =
= e^(lim(x->∞) ab) = e^ab
What if instead of taking ln on both sides you take for example base 10 log or different base log?
Then when doing L' Hopital's rule you will need to add ln(10) because of its derivative and it will get complicated
The result still holds if we replace a & b with sequences that converge to a & b, respectively. This is often encountered in statistics. See the following video. ruclips.net/video/k1BhXt9DgA4/видео.html
Why does every colour look fade besides black?
I want to be your student
i don't speak english but i love this guy, i understand better than the videos in spanish
Thank you!!!!!
Thank you so much. Your video's concept is crystal clear to me.
Why don't you utilise the definition of e....
lim x->inf (1+1/x)^x =e
lim x->inf (1+a/x)^bx
=lim x->inf [1+1/(x/a)]^[(x/a)(ba)]
={lim x->inf [1+1/(x/a)]^(x/a)}^ba
=e^ab
which is much easier....
The reason is I can show this video to my students if they have to do something like, lim as x goes to inf, (1+1/x^2)^x
This limit is 1.
www.wolframalpha.com/input/?i=limit+as+x+goes+to+inf,+(1%2B1%2Fx%5E2)%5Ex
You know, I think I see why you like the fact so much...
Great video, it was very thorough. Hopefully, this trick comes useful on my calc midterm.
yeah that's what I am telling to myself when I learn a new thing like this one...
You are an amazing teacher. Thank you
NToB36 thank you for the nice comment! You are amazing too!
Can anyone correct me if I'm wrong, but this formula should still work even if the denominator x, has a constant being added or subtracted, for example x + 2 or x - 5, cause at the end, it will eventually be zero and you will be left with (ab)/(1 + 0), which is just ab
this is awesome, thank you so much
Okay, so you used a natural logarithm to get bx out front. What if you used any other logarithm (such as log base 10) to do so? Would that not change the answer? Are you only allowed to make everything in your equation the exponent of the base of the logarithm you want to cancel out if it is a natural logarithm?
6:42 hocam silivri soğuktur L'hospitali youtube gibi herkese açık platformlarda kullanmayalım
I would not put the limit L because we do not know if the limit exists a priori..maybe this techniques gives a correct result in this case ,but in general it is not correct(for pathogenic functions)..If f(x) is the function mentioned in the video i would write: f(x)=e^ln(f(x)) and the compute the limit of ln(f(x))
Hey ... Can you do some number theory problems ?
how about if it is a number other than 1 e.g lim x-> 0 ( 2 + a/x) inside the limit (i presume the answer doesn’t involve e any more? maybe with a natural log of an integer?)
Are you sure we can't differentiate with respects to n and still use L'Hopital's rule in terms of n?
What if you take log base 10 instead of ln? would that change the answer?
How can we define asymptotes of such Functions??
Salute to you Sir! I'm a huge fan
You are great sir
Please can you make Indian language videos maths
Show that limit x approaches to + infinity then prove ln(1+a^x)/x=1
we haven't studied "l'hopital's theorem" so here is how I would do it :
I would say let t=1+a/x so then we get lim t->1 ab ln(t)/t-1=ab
Thank you for “The Fact” 👌
I like this channel. It states facts.
If I used log base 10 instead of ln I would have a different answer. Why does it have to be ln
I feel like doing the derivative of the numerator and demonstrator doesn't guarantee continuity unless the function in the numerator and denominator are the same. otherwise the delta of one will grow at a different rate and the fraction will no longer represent the same ratio. Same reason why 1/1 = 1^4/1^4 but 4/1 =/= 4^3/1^3 you can't just apply operations willy nilly on numerator and denominator.
I call this 'The Magical Limit Escape'
When all the checks, La Hospital Rule or any other rule don't lead to me any conclusion, this formula (according to you) works as a magic and leads to the exit way!!!
How would u do this without L'Hopitals ruile? :c
Thanks for the great explanations!!! BTW can you also make videos for improper integrals for finding the values of p and q in the condition of convergence?
Calisthenics Kim I have those already. ruclips.net/video/rwLkrGrugOk/видео.html
U can also go to www.blackpenredpen.com for more resources
Does that mean every continuous function can go through lim?
It is possible to solve this w/o using l'hopital right? can you please answer me how you solve this NOT using l'hopital? Thanks!
THE FACT
thanks btw, studying for math quiz tomorrow using this !
Or you could use the well known limit of ln(1+x)/x when x tends to 0. If you don't know that this is 1, you can see this this as the derivative of x\mapsto ln(1+x) in 0. Way easier than the 'hospital rule' that is something students absolutely don't understand well enough to use correctly.
My students know the LH rule.
But don't understand most probably. To the majority of student, it is like 'magic'. For proof, often they use is even if f(0)
eq 0. In any case, I think my suggestion is a bit more elementary, like you wished (do itt from scratch). Especially, using the 'hospital rule' with the indentity in the denominator seems a bit complicated for nothing :p
The limit as x goes to ∞ of (1+i/x)^(ix) = 1/e by the Fact
Shouldn't the differentiation of [ b * ln ( 1 + a/x) ] be a product rule?
Would there be a similar limit for complex z, i.e. for some sequence z_n such that for all n(in Natural numbers), there exists some N such that for all m>N, |z_m|>n, the limit lim(n->∞)(1+a/z_n)^bz_n
exists and is equal to e^ab?
And you can take the ln because you can say that there ist an x0 so that 1+a/x will be positive for all x greater than x0?
you are the best teacher!
im so grateful for you!!
You don't have to use a natural log for the calculation though, right? Would you get to the same result if you used a normal logarithm?
You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.
Roderick Llewellyn Fascinating
can you clear my doubt, that 1 to the power anything is 1 because if you keep multiplying 1 to itself its not gonna change.
so why 1 to the power infinity is not one?
and what if instead of 1 we had another constant, such as C? the result would be e^(ab/c), right?
just wondering if I replaced the natural log with other logs wouldnt that change 'The Fact'? like if I use lg it would end up 10^ab, no?
No, the answer will still be the same.
As I mentioned above: You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.
How do we know L is positive in order to use ln in it?
If I chose to take the log of both sides, wouldn't I get 10^ab as my final answer?
No because when you used the LH rule the derivative of logx isn't 1/x. /1(xln(base))
Oh okay, thanks!
lim (x to inf) (1+a/x)^bx = [t=x/a] = lim (t to inf) (1+1/t)^(abt) = [if this limit exists] (lim (t to inf) (1+1/t)^t)^ab = e^ab
any mistakes?
Thanks so much...the Fact is so useful for me...
I still don't understand why ln(lim(f(x))) = lim(ln(f(x))). You said that ln was a continuous function, which I understand, but I don't understand how that yields the swap correct algebraically.
I love your videos What is your name?
Thnku very much sir❤🙏. It has helped a lot
Great work. Do you have also a simple proof, that f(x)/g(x) equal to (d/dx f(x))/(d/dx g(x))?
Could someone explain why f(lim g(x)) only equals lim(f(g(x)) when f(x) is continuous? in this case, f(x) = lnx.
He is my legitimate god
find a and b for lim x->inf (1+a/x)^bx = pi
you can set e^(ab) = pi, then pick whatever a and b you want to make this happen.
For example, a = 1, b = ln(pi)
lim (1+i/n)^(πn) = -1
What if we had taken the log base 10 of L instead of ln in the first step. The result would be 10^ab, right?
no as the limits result would change based on how you evaluate it ? this doesnt make sense. this is because e is the only number you can differentiate in an exponential function so you have to take take ln not log 10.
@@JensenPlaysMC The function f(x)=logx is an "1-1" function. This means that x1=x2 f (x1)=f(x2). So when logx1=logx2 then and only then x1=x2. We both agree on that. In this particular situation x1=L and x2 is our limit. Why this doesn't apply here? Both logx and lnx are functions that work the same way. The difference is the base. But I don't get why e works and 10 doesn't. I'm a student, last year before university, maybe my knowledge isn't enough, but I would like to know that. Is there any link that explains it? Thanks in advance.
@@ΒαγγέληςΚωστούλας-τ2ν because when answering this question you need to use l'hospitals rule as you have an inteterminate form( differentiate numerator and denominator) in order to make sense of the limit. now. let me ask you this. when you do log 10, ( like he did to ln,e). you need you differentiate it. how do we differentiate log 10? the reason we use ln is because to differentiate any exponential function you need to use the property that e is the derivative of it self, in order to determine that the derivative of ln (1+a/x) is 1/(1+a/x) , and then use chain rule. now. you could use log 10, but then you would have to find the derivative of that which is hard to do. as if you use first princibles you see that any number differentiated apart from e^x has another string of numbers times with it
and you can express this is terms of e, much like the derivative of 2^x is (2^x * ln 2). so using any base other than e is not really good as you can express any other base in terms of e
@@JensenPlaysMC Thanks again for your reply. I think I sorted it out now. Is this a consequence of the fact that the derivative of a function is unique?
How can we know if the limit exists in the first place, cos your derivation assumed it's existence, which happens to be e. Thank you!
why could you take the derivative of the numerator and denominator? do you have a video explaining this (or another good source)?
Look up L’Hopital’s rule
but ln is not continuous everywhere