Why can we use e in these problems?

[333happy]

so after 1 year of compounding extremely many times at 100% interest rate, we get 1e. but obviously, in the real world, it's not likely that we'll literally have some money compound, say, every single second. this is my understanding: the reason why we use e is because even if we're compounding, say, every month, the number we obtain from (1+1/12 )^12 is about 2.61, and that's juuuust about e. so while it's not exactly e, it's still an okay approximation

 

[Dr.Peterson]

Are you not familiar with the concept of limit? That's what this is. The very concept of continuous compounding is a limit. It's more than a mere "speed limit", which is a number not to be exceeded; it's a number that is actually approached. Whoever wrote this was presumably writing to people who don't know about limits.

If you were compounding every second, say, you would not be compounding continuously, but approaching that. Continuous compounding means using the formula obtained as a limit, rather than merely compounding as fast as you can.

 

[JeffM]

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so after 1 year of compounding extremely many times at 100% interest rate, we get 1e. but obviously, in the real world, it's not likely that we'll literally have some money compound, say, every single second. this is my understanding: the reason why we use e is because even if we're compounding, say, every month, the number we obtain from (1+1/12 )^12 is about 2.61, and that's juuuust about e. so while it's not exactly e, it's still an okay approximation

In the US, banks used to advertise continuous compounding. We used a formula involving e rather than computing interest frequently.

 

[pka]

This is always a fascinating question. But as Prof. Peterson pointed out unless one understands limits this may not help.
This from Berresford&Rockford's Applied Calculus: consider a sum of money \(P\) deposited at \(16\%\) which is compounded quarterly for one year. We would get back \(P\left(1+\dfrac{.16}{4}\right)^4=P(1+.0.04)^4\approx 1.16986\times P\).
If it were compounded monthly \(P\left(1+\dfrac{.16}{12}\right)^{12}\approx 1.17227\times P\) a bit more. But not much!
Suppose it were compounded daily \(P\left(1+\dfrac{.16}{365}\right)^{365}\approx 1.17347\times P\) a bit more. But not much!
Let's now look at continuous compounding. To do the we need the limit \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \dfrac{r}{n}} \right)^n} = {e^r}\) where \(r\) is the interest rate. We let the number of compoundings increase without bound.
Let's use \(0.16\): \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \dfrac{0.16}{n}} \right)^n\times P} = e^{0.16}\approx 1.173511\times P\)
That is only a bit more, but the idea is appealing none-the-less.