@@anglaismoyen "The Fact" is: lim n->∞ (1 + a/n)^bn = e^ab It's derived from the limit definition of e: lim n->∞ (1 + 1/n)^n = e It's also the limit of the formula for Compound Interest so you can search for that too.
It is possible to notice that 1/sin x when x approaching zero as the same size of x (x->0 => sen x / x = 1) and then use the fact that y->0 => (1+y)^(1/y) -> e (y=-3x)
I saw this videos thumbnail and did the problem. Just watched the video and I went through the same exact steps you did :0 the ln/e^lnx trick is so so useful
Taking the limit of a continuous function is the same as taking the function of the limit. For example x^2 is a continuous function so we can say: limit x^2 × --> 0 = (limit x ) ^2 x---> 0 Where we put the square outside of the limit for the same reason.
I'm unhappy with what happens at 2:28: "Because ln is a continuous function, we can....". No it's not. You are already assuming that the limit is positive. You can do this as a tentative assumption that you have to check at the end: exp(-3) > 0. BTW: I think one could do this limit by inspection using sin(x)~x near 0, and (in your words) "The Fact".
The limit is an exponent, the base is positive because for x being a number really close to zero ( negative / positive ) 1 - 3x will be a strictly positive number
@@Michel-dx1bn (-1)^x has a really strange domain: Obviously, that function is well-defined for all integers. One could also argue that it is defined for all fractions with odd denominators, since one could argue that e. g. (-1)^(1/3) = third root of -1 = -1. (As far as I know, that's not usual, but some math books allow that.) So the graph consists of infinitely many points. In the first case with only integers, these points are well-separated from each other; in the second case, with some fractions allowed, the points lie dense, but nevertheless the line is not continuous, but still has infinitely many gaps. And in the latter case, you get a line which is wildly oscillating up- and downwards, I don't think one could actually draw that graph in a sensible way. Coincidentally, I just posed this question to my own students only about one week ago. :D But it was a facultative exercise and I strongly suspect that none of them have actually tried to solve this. :/
I wonder if there's a way to make this valid/rigorous: lim x->0 (1-3x)^(cscx) lim x->0 (1-3x)^(x/xsinx) lim x->0 ((1-3/(1/x))^(1/x))^(x/sinx) lim x->0 (e^-3)^(1)=e^-3
Im probably just bored but how would one prove that (a+b)(c+d)=ac+ad+bc+bd? It seems obvious since we are used to multiplying the terms in a cross to get correct result but that's just a trick and not really a proof why it holds
yes. Denote Your Expression as YE. You can rewrite it as exp{ln{YE}}. take the exponent from inside of logarithm to the front, apply product rule for limits and go back. Hard to explain it better in that comment section. Basically, Your approach is correct ;)
I guess there's a way to solve this without using differentiation (in case the student didn't get to this theme and there's only limits). 1 to the power of infinity just yells "Use (1+x)^(1/x) limit". That leads to the e to the power of -3x/sin(x) and the limit of x/sin(x) is 1. No l'Hopital, so it's possible to be solved by the 1st year student.
Doesn't evaluating x/sin(x) = 1 as x goes to zero still require L'Hopital? Sure, you could write out sin as a polynomial (~x), but understanding Taylor Series expansion still requires knowledge of derivatives by definition.
@@Max_Miles It doesn't. It's one of the fundamental limits (the second one is the (1+1/x)^x limit) and can be proven without Taylor, l'Hopital or any differentiation.
Very nice idea to rewrite such that l’hospital’s rule can be used. I saw a simplified proof of this rule and it made sense. Maybe he came up with it that way? It’s a beautiful tool
🎉🎉I don't know about teaching in your country but in india we are familiar with 1^infinity by just using one formula for each problems of form 1^infinity .If you don't know about it I would like to share 😊😊 general form : lim x-->a [f(x)]^(g(x)) Where f(a)-->1 g(a)-->infinity Then value of limit at a is e^(g(x)•(1-f(x))) 😊😊 Proof is here (f(x))^g(x)=(1+(f(x)-1))^g(x) =(1+(f(x)-1))^[g(x)•(f(x)-1)]/f(x)-1 For avoiding complexity during typing I write i=f(x)-1 OK And g(x)•(f(x)-1)=gi Then (1+i)^(gi/i) And ((1+i)^(1/i))^gi we know limit as I tence to 0 (1+i)^(-i)=e Hence game finished e^gi And gi is already mentioned 😊😊😊😊
Found this Here’s how it works: Start with the expression (f(x)^{g(x)}). Rewrite it as (\exp[\log(f(x)) \cdot g(x)]), where (\exp) represents the exponential function and (\log) is the natural logarithm. Next, express the logarithm as (\log(f(x)) = f(x) - 1). This step relies on the fact that (\lim_{{u \to 0}} \frac{{e^u - 1}}{u} = 1). Now we have (\exp[(f(x) - 1) \cdot g(x)]). Finally, evaluate the limit: (\lim_{{x \to a}} f(x)^{g(x)} = \exp[\lim_{{x \to a}} (f(x) - 1) \cdot g(x)]).
I did it this way: lim x -> 0 (1 - 3x)^csc(x) = lim x -> 0 (1 - 3x)^(1/sin(x))= = lim x -> 0 (1 - 3x) ^ (x/sin(x) * 1/x). Thanks to a famous limit we know that lim x -> 0 sin(x)/x = lim x -> 0 x/sin(x) = 1. So we get: lim x -> 0 (1 - 3x) ^ 1/x. We could have achieved this result also with MacLaurin of sin(x) which (for x -> 0) is approximately equal to x. Now we can write the limit as: lim x -> 0 (1 - 3x) ^ (1/-3x * (-3)). Now we have a limit in the form of lim x -> 0 (1 + f(x))^1/f(x) = e Thus, lim x -> 0 (1 - 3x) ^ (1/-3x * (-3)) = [ lim x -> 0 (1 - 3x) ^ (1/-3x) ] ^ (-3) = e^-3
No, in L'H rule, you take the derivative of the numerator and denominator separately (provided that the original limit is indeterminate). It's not the same as taking the derivative of the limit
Thinking about it again, it is a little silly to talk about derivative of limit. Limit itself will be a constant (or not defined). It makes more sense to take limit of derivative. Sorry for chaos, I was interpreting an extra layer of differentiating the entire expression.
No, the limit of a fraction f/g is equal to the limit of the fraction f’ / g’, only in the case when the limits of f and g are both 0 or both infinity or negative infinity. Look up L’Hospitals Rule. BPRP has done lots of videos on it (and you can also find plenty of discussion on it in any calculus textbook).
Alternative Solution: lim x->0 (1+(-3)*1*x)^cscx = lim x->0 (1+(-3)*(1/(x/sinx))*x)^cscx (Because sinx/x goes to 1 when x goes 0) =lim x->0 (1+(-3)*(1/(cscx)))^cscx So lets say cscx=t, when x goes 0, t goes infinity. So new limit is; lim t->∞ (1+(-3)*t)^1*t = e^(-3)
Yes, in this case it is basically the same as saying, lim (1-3x)^csc(x) = lim e^ln((1-3x)^csc(x)) and then you use the continuity of the e-function to switch e^ and the limit and one can then work out the new limit using L'Hospital's rule. If the limit does not exist, imo the best option to show that (especially when you have sth with sin, cos, etc.) is to find two different sequences x_n, y_n convergent to 0, such that for example sin(x_n) = 1 and sin (y_n) = 0 for all n, or whatever works in the given situation, such that you get different values for the limit.
They doing construction!!!! th-cam.com/video/dX7iGKJsIhY/w-d-xo.html
Why you can’t say 1^inf=1
th-cam.com/video/1ebqYv1DGbI/w-d-xo.html
Quick and dirty method. as x->0 sin(x)=x
substituting in we get (1-3x)^1/x. After substituting t=1/x, then by "The Fact" this is equal to e^-3
I was thinking about a limited Taylor series so yeah ^^
I've heard bprp reference 'the fact'. Do you know the proper name for it so I can read more about it?
@@anglaismoyen "The Fact" is:
lim n->∞ (1 + a/n)^bn = e^ab
It's derived from the limit definition of e:
lim n->∞ (1 + 1/n)^n = e
It's also the limit of the formula for Compound Interest so you can search for that too.
that edit to erase the white board was very smooth!
graphing this, i came across a very near coincidence: e^-3 is very very close to 1/20
It‘s very close to 1/20.
@@Bayerwaldler oh sorry, fixed it
Yep, e^3 is pretty close to 20. Matches up with the first digit after the dot
e^π - π ≈ 20
It is possible to notice that 1/sin x when x approaching zero as the same size of x (x->0 => sen x / x = 1) and then use the fact that y->0 => (1+y)^(1/y) -> e (y=-3x)
Damn, you really do get to know something new everyday!
I saw this videos thumbnail and did the problem. Just watched the video and I went through the same exact steps you did :0 the ln/e^lnx trick is so so useful
lyrics:
I'm
I'm not sure how to ta-
2:24 I don't understand this passage: why can you write lim in front of ln?
Taking the limit of a continuous function is the same as taking the function of the limit.
For example x^2 is a continuous function so we can say:
limit x^2
× --> 0
= (limit x ) ^2
x---> 0
Where we put the square outside of the limit for the same reason.
@@TSSPDarkStar thank you😉
after learning limits and derivatives I finally understand the calculus he does in his videos after years of watching
I'm unhappy with what happens at 2:28: "Because ln is a continuous function, we can....". No it's not. You are already assuming that the limit is positive. You can do this as a tentative assumption that you have to check at the end: exp(-3) > 0.
BTW: I think one could do this limit by inspection using sin(x)~x near 0, and (in your words) "The Fact".
The limit is an exponent, the base is positive because for x being a number really close to zero ( negative / positive )
1 - 3x will be a strictly positive number
Exponential functions are always positive, so the limit has to be positive. No assumptions
@@davidcroft95 if the base is positive. although i am note sure how the graph of something like (-1)^x would look like
@@Michel-dx1bn if it's negative it's bad defined: the function doesn't not exist (in that interval, that is) in the first place
@@Michel-dx1bn (-1)^x has a really strange domain: Obviously, that function is well-defined for all integers. One could also argue that it is defined for all fractions with odd denominators, since one could argue that e. g. (-1)^(1/3) = third root of -1 = -1. (As far as I know, that's not usual, but some math books allow that.)
So the graph consists of infinitely many points. In the first case with only integers, these points are well-separated from each other; in the second case, with some fractions allowed, the points lie dense, but nevertheless the line is not continuous, but still has infinitely many gaps. And in the latter case, you get a line which is wildly oscillating up- and downwards, I don't think one could actually draw that graph in a sensible way.
Coincidentally, I just posed this question to my own students only about one week ago. :D But it was a facultative exercise and I strongly suspect that none of them have actually tried to solve this. :/
I wonder if there's a way to make this valid/rigorous:
lim x->0 (1-3x)^(cscx)
lim x->0 (1-3x)^(x/xsinx)
lim x->0 ((1-3/(1/x))^(1/x))^(x/sinx)
lim x->0 (e^-3)^(1)=e^-3
Im probably just bored but how would one prove that (a+b)(c+d)=ac+ad+bc+bd? It seems obvious since we are used to multiplying the terms in a cross to get correct result but that's just a trick and not really a proof why it holds
yes. Denote Your Expression as YE. You can rewrite it as exp{ln{YE}}. take the exponent from inside of logarithm to the front, apply product rule for limits and go back. Hard to explain it better in that comment section. Basically, Your approach is correct ;)
I guess there's a way to solve this without using differentiation (in case the student didn't get to this theme and there's only limits).
1 to the power of infinity just yells "Use (1+x)^(1/x) limit". That leads to the e to the power of -3x/sin(x) and the limit of x/sin(x) is 1. No l'Hopital, so it's possible to be solved by the 1st year student.
Doesn't evaluating x/sin(x) = 1 as x goes to zero still require L'Hopital? Sure, you could write out sin as a polynomial (~x), but understanding Taylor Series expansion still requires knowledge of derivatives by definition.
@@Max_Miles It doesn't. It's one of the fundamental limits (the second one is the (1+1/x)^x limit) and can be proven without Taylor, l'Hopital or any differentiation.
@@SURok695 huh, neat! Thanks for clearing that up.
Very nice idea to rewrite such that l’hospital’s rule can be used. I saw a simplified proof of this rule and it made sense. Maybe he came up with it that way? It’s a beautiful tool
This was super clear thank you I didn’t know how to solve that
🎉🎉I don't know about teaching in your country but in india we are familiar with 1^infinity by just using one formula for each problems of form 1^infinity .If you don't know about it I would like to share 😊😊
general form :
lim x-->a [f(x)]^(g(x))
Where
f(a)-->1
g(a)-->infinity
Then value of limit at a is
e^(g(x)•(1-f(x))) 😊😊
Proof is here
(f(x))^g(x)=(1+(f(x)-1))^g(x)
=(1+(f(x)-1))^[g(x)•(f(x)-1)]/f(x)-1
For avoiding complexity during typing I write i=f(x)-1 OK
And g(x)•(f(x)-1)=gi
Then
(1+i)^(gi/i)
And
((1+i)^(1/i))^gi
we know limit as I tence to 0
(1+i)^(-i)=e
Hence game finished
e^gi
And gi is already mentioned 😊😊😊😊
Found this
Here’s how it works:
Start with the expression (f(x)^{g(x)}).
Rewrite it as (\exp[\log(f(x)) \cdot g(x)]), where (\exp) represents the exponential function and (\log) is the natural logarithm.
Next, express the logarithm as (\log(f(x)) = f(x) - 1). This step relies on the fact that (\lim_{{u \to 0}} \frac{{e^u - 1}}{u} = 1).
Now we have (\exp[(f(x) - 1) \cdot g(x)]).
Finally, evaluate the limit: (\lim_{{x \to a}} f(x)^{g(x)} = \exp[\lim_{{x \to a}} (f(x) - 1) \cdot g(x)]).
I did it this way:
lim x -> 0 (1 - 3x)^csc(x) = lim x -> 0 (1 - 3x)^(1/sin(x))=
= lim x -> 0 (1 - 3x) ^ (x/sin(x) * 1/x).
Thanks to a famous limit we know that lim x -> 0 sin(x)/x = lim x -> 0 x/sin(x) = 1. So we get:
lim x -> 0 (1 - 3x) ^ 1/x.
We could have achieved this result also with MacLaurin of sin(x) which (for x -> 0) is approximately equal to x.
Now we can write the limit as:
lim x -> 0 (1 - 3x) ^ (1/-3x * (-3)). Now we have a limit in the form of lim x -> 0 (1 + f(x))^1/f(x) = e
Thus, lim x -> 0 (1 - 3x) ^ (1/-3x * (-3)) = [ lim x -> 0 (1 - 3x) ^ (1/-3x) ] ^ (-3) = e^-3
I thought you were going to take derivative of the limit, unsure how that can be done. Can we simply swap derivative and limit?
No, in L'H rule, you take the derivative of the numerator and denominator separately (provided that the original limit is indeterminate).
It's not the same as taking the derivative of the limit
Thinking about it again, it is a little silly to talk about derivative of limit. Limit itself will be a constant (or not defined). It makes more sense to take limit of derivative.
Sorry for chaos, I was interpreting an extra layer of differentiating the entire expression.
You can take the derivative of a limit. The derivative itself is defined by using limits. This is exactly what finding "the second derivative" is
How do you prove: 😯 limit as x→ ∞ of (x!)²(e/x)^(2x)/(2x) = π
Wait, the limit of a function is the same as the limit of its derivative?
No, the limit of a fraction f/g is equal to the limit of the fraction f’ / g’, only in the case when the limits of f and g are both 0 or both infinity or negative infinity. Look up L’Hospitals Rule. BPRP has done lots of videos on it (and you can also find plenty of discussion on it in any calculus textbook).
Alternative Solution:
lim x->0 (1+(-3)*1*x)^cscx = lim x->0 (1+(-3)*(1/(x/sinx))*x)^cscx
(Because sinx/x goes to 1 when x goes 0)
=lim x->0 (1+(-3)*(1/(cscx)))^cscx
So lets say cscx=t, when x goes 0, t goes infinity. So new limit is;
lim t->∞ (1+(-3)*t)^1*t = e^(-3)
Is it legitimate to set the limit equal to *something* if it doesn't exist? In that case, what would be the procedure?
Yes, in this case it is basically the same as saying, lim (1-3x)^csc(x) = lim e^ln((1-3x)^csc(x)) and then you use the continuity of the e-function to switch e^ and the limit and one can then work out the new limit using L'Hospital's rule. If the limit does not exist, imo the best option to show that (especially when you have sth with sin, cos, etc.) is to find two different sequences x_n, y_n convergent to 0, such that for example sin(x_n) = 1 and sin (y_n) = 0 for all n, or whatever works in the given situation, such that you get different values for the limit.
Essentially that's always done when something tends to infinity: No limit exists, but nevertheless one says that the limit is infinity.
What about using (1+1/an)^an?
I accidentally thought it meant I was going to derive the pure function, will still do it later though
Why you can’t say 1^inf=1
th-cam.com/video/1ebqYv1DGbI/w-d-xo.html
peak