A COUNTER-INTUITIVE CALCULUS LIMIT

We know the limit of (1+1/n)^n goes to the Euler's number e as n goes to infinity. However, we need to be extra careful when we are evaluating the limit of n((1+1/n)^n-e) as n goes to infinity. This is a great L'Hospital's rule limit question to challenge calculus 1 and calculus 2 students.

#maths #calculus #blackpenredpen #mathforfun

[josephl]

Casually doing math while holding a thermal detonator.

[AlRoderick]

It's clever for you to use the blue pen for "I did this in a different video" because it makes me think your whiteboard has hyperlinks. When I go watch that video and come back will that part turn purple?

[lswcs]

The "please pause the video and first try for yourself" is always so motivating :)
I was lying in my bed and this motivated me to get up and try it :)

[altacc]

It helps when you keep in mind that the arithmetic rules for limits of sequences only apply if all sequences used actually converge :)

[masongatz]

Weird things can happen with infinity. This was one of my favorite videos so far!

[timtimtimmaah]

That it's negative this makes sense considering that limit definition of e approaches e from the left of e.

[krukowstudios]

I’d say 0. “en” means “one” in Danish, so it is 1 - 1.

[modemanslutning]

For the "intuitive" explanation:

The 2nd part, x*e, is like a typical linear function: y=k*x, with the constant being "e". This of course grows linearly.

The 1st part, x*(1+1/x)^x, will eventually reach a similar value, the latter part eventually becoming "e", so this term will also become a linear equation essentially, x*e. But it takes a while for it to grow and get closer to the value that is "e".

Thus x*(1+1/x)^x is "lagging behind" the other function. After a while it will stabilize though when becoming more linear. The difference between the two functions caused by that lag will thus be e/2 (the minus sign in the limit depends on which of the functions we put first).

If you want to see this yourself, plot the two graphs:
y(x) = x*e
g(x) = x*(1+1/x)^x

And measure the height difference for some relatively high value of x. Should be about e/2 = 1.36.

[tollboothjason]

That was some fast writing at 6:44 and 6:48! 😁

[paul_w]

Using ln(1+1/n)=1/n-1/2n²+o(1/n²) and e^(1/n)=1+1/n+o(1/n) you can solve that easily and quickly without using l'hôpital rule and without these weird derivatives.
Taylor series are so useful to find limits...
You make this looks really hard when it really isn't.

[restcure]

Yesterday the beard ... today the plaid flannel shirt ... tomorrow - full lumberjack?

edit: re-watching after three weeks (forgot method - whoops) I felt the desire to add: time to remember it was "x", not "n": about 27 seconds - been there!

[FlyNavy]

I was having a crisis thinking Q1 might not be 0 somehow lol

[ryanchatterjee]

This guy's living the dream: beautiful wife, genius son, wears stylish clothes, makes math videos for a living.

[ramez]

One of the best math channels out there - deserves more subscribers! Keep up the good work!

[hugolaurent]

You can also use the Taylor expansion twice, with ln (because (1+1/n)^n = e^nln(1+1/n))
And again with exp (you factorizes by ne : ne*(e^-1/2n - 1) and the Taylor expansion of exp(-1/2n) gives us 1-1/2n so finally ne*(-1/2n) which is equal to - e/2
I find this quicker ;)

[xavierplatiau]

Believe it or not, I did it in my head, using limited developments of ln(1+x) as x goes to 0 and of exp(x).
Did it like (1+1/n)^n - e = exp(n ln(1+1/n) ) - e ~ exp(n(1/n - 1/2n^2)) - e = exp(1 - 1/2n) - e = e(exp(-1/2n) - 1) ~ -e/2n
And then multiplied by n it leads to -e/2.

[dugong]

In the process of looking more into your video about generating x and y such that x^y=y^x, I found that (1+1/x)^x and (1+1/x)^(x+1) generates such a pair (directly derived from your method using t). Since these pairs always bracket e, and are almost equidistant from e when they are close to e, a better estimate for e is (1+1/n)^(n+1/2). If you substitute this better estimate for e in your expression and then factor out (1+1/n)^n (and call that e like you did in this video) and then find the limit of the remaining factor, also using the fact that for large n, sqrt(n(n+1)) is about n + 1/2, that limit is quickly found to be -1/2, giving the correct answer -e/2 (not by such a rigorous process, but you might be able to justify it better than I could). No LH rule or derivatives involved, just algebra.

Edit: I went through the LH rule procedure for the limit of ((1+1/x)^(x+1/2)*x - e*x) and unless I made an algebra error, this limit is 0. The polynomials in numerator and denominator are one degree higher, but the numerator still cancels out to 1. [Just checked it on Wolfram Alpha, the limit is 0.]

[PackSciences]

That moment when you forget a minus sign

[gillrowley]

Stop blowing my mind!! Actually, keep going. Fascinating videos that bring back some knowledge I haven't used in a long time.

[FuhrerShattercore]

The answer makes perfect sense since that would be the accumulation of errors between the limit and the value e. I expected it on the value becoming negative but arriving on exactly -e/2 was pretty amazing