Casually doing math while holding a thermal detonator.

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?

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

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

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 :)

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

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

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

That moment when you forget a minus sign

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

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!

70% of the time i was looking at his beard

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.

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 ;)

Just great! Thanks from Switzerland.

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.

4:30am. Good morning, time for BlackpenRedpen

Another approach is to do the binomial expansion of (1+1/n)^n and then subtract the series expansion of e. Multiply the difference by n and the answer falls out. No l’hopitals needed.

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

You and Peyam are the best