Second last step: you need to use "continuity" of ln(x). (Because:with still finite x/h the term in the ln is not yet 'e'). BTW: I love the way you are presenting this explaining every step (in a calm and friendly way). Most teachers try to do it in a hurry. That is the main reason, why some students get annoyed with math.
Hello sir, i'm seeing that we always using definition of e with limit while proving all of these formulas but is it actually possible for you to explain or prove how or why limit n->infinity (1+1/n)^n is equal to euler's number? Did you record a video about this or would you talk about that in another video if it's possible? Thanks
If you define ln(x) as the integral from 1 to x of 1/u du, you have no problems computing its derivative. It is equal to 1/x by definition of the function.
just a suggestion 😁 It might be too late to point out but, you could have started by first proving that ln(x) is indeed differentiable by Left and Right hand derivative then go on to find it.
This works because lim x-> -inf (1+1/x)^x is also equal to e, but I don't know any proofs without using the derivative of ln(x) (this would be circular reasoning).
@@xavierwainwright8799 take lim x-> -inf (1 + 1/x)^x substitute w = -x, so as x -> inf w -> inf = lim w -> inf (1 - 1/w)^(-w) = lim w -> inf e^ ln((1 - 1/w)^(-w)) bring -w out front with log rules = lim w -> inf e^( -w * ln(1 - 1/w) ) rewrite the subtraction inside the natural log = lim w -> inf e^( -w * ln((w - 1)/w) ) rewrite division inside ln as subtraction of lns = lim w -> inf e^( -w * ( ln(w - 1) - ln(w) ) ) use the negative sign on w in the exponent to switch the order of subtraction = lim w -> inf e^( w * ( ln(w) - ln(w - 1) ) ) recombine logs and bring the w inside as an exponent = lim w -> inf e^ln( (w / (w - 1)) ^ w ) cancel the exponential and log = lim w -> inf (w / (w - 1))^w substitute w = n + 1, so as w -> inf n also -> inf = lim n -> inf ((n + 1)/n))^(n + 1) take out the base of the n+1 exponent to get rid of the 1 = lim n-> inf ((n+1)/n)^n * (n+1)/n write limit of product as product of limits = lim n-> inf ((n + 1)/n)^n * lim n-> inf (n + 1)/n first limit is the normal form of the limit for e, second limit is easily calculated to be 1 = e
Second last step: you need to use "continuity" of ln(x). (Because:with still finite x/h the term in the ln is not yet 'e').
BTW: I love the way you are presenting this explaining every step (in a calm and friendly way). Most teachers try to do it in a hurry. That is the main reason, why some students get annoyed with math.
I never thought of this! Good job!
the enthusiasm made this really enjoyable to watch, great job
Beautiful! 1st time seeing this truly from first principles!
This guy turns math into magic
Very good explanation ❤
I really like your classes, thank you for your hard work! 😃
Very good. Thanks 👍
You are AWESOME!!
Great job
Love your videos
Solid videos!
You are the best
It's very useful to understand the inderivatived integral of dx/x sir!
Use grouping symbols: (dx)/x
You are a great man
8:58 as h goes to 0, "n" or x/h goes to infinity. Thus, lim as h goes to 0 = lim n goes to infinity.
Hello sir, i'm seeing that we always using definition of e with limit while proving all of these formulas but is it actually possible for you to explain or prove how or why limit n->infinity (1+1/n)^n is equal to euler's number? Did you record a video about this or would you talk about that in another video if it's possible? Thanks
Love this
If you define ln(x) as the integral from 1 to x of 1/u du, you have no problems computing its derivative. It is equal to 1/x by definition of the function.
just a suggestion 😁
It might be too late to point out but, you could have started by first proving that ln(x) is indeed differentiable by Left and Right hand derivative then go on to find it.
😍😍👌👌✊✊👍👍
5:31, as h goes to zero wouldn't x/h go to either positive infinity or negative infinity?
This works because lim x-> -inf (1+1/x)^x is also equal to e, but I don't know any proofs without using the derivative of ln(x) (this would be circular reasoning).
@@xavierwainwright8799
take lim x-> -inf (1 + 1/x)^x
substitute w = -x, so as x -> inf w -> inf
= lim w -> inf (1 - 1/w)^(-w)
= lim w -> inf e^ ln((1 - 1/w)^(-w))
bring -w out front with log rules
= lim w -> inf e^( -w * ln(1 - 1/w) )
rewrite the subtraction inside the natural log
= lim w -> inf e^( -w * ln((w - 1)/w) )
rewrite division inside ln as subtraction of lns
= lim w -> inf e^( -w * ( ln(w - 1) - ln(w) ) )
use the negative sign on w in the exponent to switch the order of subtraction
= lim w -> inf e^( w * ( ln(w) - ln(w - 1) ) )
recombine logs and bring the w inside as an exponent
= lim w -> inf e^ln( (w / (w - 1)) ^ w )
cancel the exponential and log
= lim w -> inf (w / (w - 1))^w
substitute w = n + 1, so as w -> inf n also -> inf
= lim n -> inf ((n + 1)/n))^(n + 1)
take out the base of the n+1 exponent to get rid of the 1
= lim n-> inf ((n+1)/n)^n * (n+1)/n
write limit of product as product of limits
= lim n-> inf ((n + 1)/n)^n * lim n-> inf (n + 1)/n
first limit is the normal form of the limit for e, second limit is easily calculated to be 1
= e
Recall that the domain of ln (x) is **positive real numbers only** hence x/h is a positive real number x divided by a quantity h tending to zero
Awesome
But n is integer while x/h is real number?
BRAVO
(1 + 1/(x/h))^(x/h) is not equal e
Always the best teacher.....what about 1/x
Thanks
He missed step when he used fact that ln is continuous
You are the best
Thank you for your kind words.