1:55 A minor tweak, but once you have limit as x goes to infinity of ln(1+ 1/x) / (1/x) , you can simplify the next step a bit by substituting z = 1/x . Now you are looking at limit at z approaches 0+ of ln(1+z) / z . Using L’Hopital as in the video, the derivatives are now slightly simpler and you get just the limit at z approaches 0+ of 1 / (1 + z) which is 1. So ln y = 1, and y = e.
You cant use ln in your proof since ln is based on e and its derivative is based on the results of this theorem. The actual proof is complicated and it just proves the limit exists and is between 2 and 3, than it is calculated with taylor approximation
@@blackapple62 well, in more advanced mathematics the natural logarithm log(x) is often defined as the integral from 1 to x of dt/t. Then you prove that it's also the inverse function of exp(x).
What I'd be interested in knowing is what happens if 1/x is replaced by 2/x, or 3/x. Since the given formula converges and one factor is being multiplied by a constant, the new formula should also converge.
Basically, he assigned "y" to the equation as a variable. Therefore, we assumed that "y" equaled the whole equation. In the case of (ln y = lim), we multiplied both sides with "ln" (excluding the limit). Later, he explained it in more detail.
I am getting stuck like why are we cancelling -1/x^2 won't those two be equal to 0 and we cannot cancel 0 terms like that Please correct me if I am wrong
You can cancel equal non-zero factors from the numerator and denominator. They may be small but mustn't be zero. You can't do the cancellation if both are zero. This is the main idea in limit evaluation.
ln == log base e , and e is defined by this limit, so nothing was achieved other than showing consistency - there is no need to prove anything, as the definition of e itself is the limit
@@ntlake that’s the exact same thing?? Using n,x,α,γ,.. whatever letter u want doesn’t matter as it is a dummy variable And if u mean n as in natural number or discrete limit, it doesn’t matter as the limiting function is continuous anyways, the discrete and continuous limits must agree
@@adw1z yeah, but no. The definition of e is the limit of a sequence, then it being equal to the limit of the function is something you have to prove. Anyway, I can define e without using this limit if you want.
@@adw1z sorry, missed the notification. the fact that the proof is trivial doesn't change the fact that it isn't its definition. Anyway the other definition is e = 1/0! + 1/1! + 1/2! + 1/3! + ...
Excellent class, I can 100% digest the outcome of e, your patience and derive step by step is unbeatable.
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1:55 A minor tweak, but once you have limit as x goes to infinity of ln(1+ 1/x) / (1/x) , you can simplify the next step a bit by substituting z = 1/x . Now you are looking at limit at z approaches 0+ of ln(1+z) / z . Using L’Hopital as in the video, the derivatives are now slightly simpler and you get just the limit at z approaches 0+ of 1 / (1 + z) which is 1. So ln y = 1, and y = e.
You cant use ln in your proof since ln is based on e and its derivative is based on the results of this theorem.
The actual proof is complicated and it just proves the limit exists and is between 2 and 3, than it is calculated with taylor approximation
Not necessarily. You can define ln(x) without using e at all.
@@ntlake How exactly can you do this? Using the natural logarithm quite literally means you're using a logarithm of base e.
@@blackapple62 well, in more advanced mathematics the natural logarithm log(x) is often defined as the integral from 1 to x of dt/t. Then you prove that it's also the inverse function of exp(x).
@@ntlakeexactly
you can do it by figuring it out as an area function or antiderivative of 1/x
You can not use natural logarithm as E is to be proved. Dhanyavad
you can if you define lnx as the integral from 1 to x of 1/t dt, and define e as the point where ln(x)=1.
hello, thank you for your video, but I am confused about how you canceled the -1/x^2.
Awesome explanation!
What I'd be interested in knowing is what happens if 1/x is replaced by 2/x, or 3/x. Since the given formula converges and one factor is being multiplied by a constant, the new formula should also converge.
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شكرا لكم على المجهودات.
🙏🙏
but isn't this limit the definition of e? or are you using the infinite polynomial definition of e^x
It's indeed the definition
There is only one definition lim(1+1/x)^x. It is a continuous compounding
I understood all, but at the beguining, why there is (let y) what meaning ?, and what ln y = lim.....
Basically, he assigned "y" to the equation as a variable. Therefore, we assumed that "y" equaled the whole equation. In the case of (ln y = lim), we multiplied both sides with "ln" (excluding the limit). Later, he explained it in more detail.
@@lubanlatif3713 ok thanks !
I just discovered your channel. You are doing a wonderful job. Do you have a Whatsapp or telegram group for Jss1 student's mathematics
I appreciate it 🙏
We only have a group on Facebook
Excellent
Idk, there was a n easier method imo, where by using log properties and mclaurin you not only simplify the process but also save up time.
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Very cool thank you
Thank you
Binomial expansion formula need to be used because e is to be proved.
I am getting stuck like why are we cancelling -1/x^2 won't those two be equal to 0 and we cannot cancel 0 terms like that
Please correct me if I am wrong
You can cancel equal non-zero factors from the numerator and denominator. They may be small but mustn't be zero. You can't do the cancellation if both are zero. This is the main idea in limit evaluation.
logy = 1 and then y = 10 or lim xlog(1+1/x) = 10 when x is very bigger
What will happen if base is 10? ( Then we have log 10 = 1 )
ln == log base e , and e is defined by this limit, so nothing was achieved other than showing consistency - there is no need to prove anything, as the definition of e itself is the limit
Nope. e isn't defined by this limit, it's defined by the limit as n approaches infinity of (1+1/n)ⁿ
@@ntlake that’s the exact same thing?? Using n,x,α,γ,.. whatever letter u want doesn’t matter as it is a dummy variable
And if u mean n as in natural number or discrete limit, it doesn’t matter as the limiting function is continuous anyways, the discrete and continuous limits must agree
@@adw1z yeah, but no. The definition of e is the limit of a sequence, then it being equal to the limit of the function is something you have to prove.
Anyway, I can define e without using this limit if you want.
@@ntlake the proof follows trivially directly from the definition of continuity of a function, but I’d be interested in seeing your other definition
@@adw1z sorry, missed the notification. the fact that the proof is trivial doesn't change the fact that it isn't its definition.
Anyway the other definition is e = 1/0! + 1/1! + 1/2! + 1/3! + ...
awesome