$A question about the mathematical constant e.$

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$A question about the mathematical constant e.$

I understand that when we write e = lim⁡ n→∞ (1+1/n)^n, we are identifying the number that this expression converges to as n becomes very large. However, I also know that lim⁡ n→∞(1/n)= lim⁡ n→∞(100/n )=0. Given this, why doesn't the same logic apply to e, such that lim⁡ n→∞ (1+1/n)^n = lim n→∞ (1+100/n)^n = e?

I've seen the proof that shows:

Starting from the definition of e:

e = lim⁡ n→∞ (1+1/n)n

Exponentiating both sides by k:

e^k = (lim⁡ n→∞ (1+1/n)^n)^k

Using properties of limits and exponents:

e^k = lim⁡ n→∞ ((1+1/n)^n)^k

Simplifying the expression inside the limit:

((1+1/n)^n)^k = (1+1/n)^nk

Substituting m=nk:

n=m/k

(1+1/n)^n=(1+k/m)^m

Applying this to k=100:

e^100 = lim n→∞ (1+100/n)^n

But I want a more intuitive understanding of why lim⁡ n→∞ (1+1/n)^n = e while lim⁡ n→∞ (1+100/n)^n = e^100, even though 1/n and 100/n both approach zero as n tends to infinity.

Calculus

Mathgnome

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The reason is that not all small numbers are the same. In other words, there are different rates of convergence. Here $1/n$ and $100/n$ both converge to 0, but at different rates. In fact $100/n$ is 100 times larger than $1/n$, although both are very small numbers when $n$ is large. When we just look at their limits we do not see this difference, they both converge to 0. Equivalently both $$(1+\frac{1}{n}),\quad (1+\frac{100}{n})$$ converge to 1 as $n\to \infty$, although $1+\frac{1}{n}$ converges to 1 faster, but we do not see this difference when we just look at the limit. However when we raise them to power $n$, which is a very large number, this difference becomes noticable. So the difference of the above two expressions, i.e. $\frac{99}{n}$, goes to 0 and is very small, but even this small difference can become important when we raise the expressions to the very large power $n$. Using the binomial theorem we can also see $$ (1+\frac{100}{n})^n = (1+\frac{1}{n} +\frac{99}{n})^n \\ \; \\=(1+\frac{1}{n})^n +n(1+\frac{1}{n})^{n-1}(\frac{99}{n})+ \dots +n(1+\frac{1}{n})(\frac{99}{n})^{n-1}+ (\frac{99}{n})^n$$ And the difference $n(1+\frac{1}{n})^{n-1}(\frac{99}{n})+ \dots +n(1+\frac{1}{n})(\frac{99}{n})^{n-1}+ (\frac{99}{n})^n$ is so large as $n\to \infty$ that changes the limit from $e$ to $e^{100}$.

Martin

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Mathgnome

0

When you say "In fact 100/n is 100 times larger than 1/n, although both are very small numbers" Does that mean realistically 100/n and 1/n are different numbers, but the difference between them is infinitesimal allowing the difference to be ignored.
- Martin
  
  +1
  
  Yes, the difference is very small and can be ignored (before raising to the power of n of course).
Mathgnome

0

Thanks so much for the explanation it has really helped me out.
- Martin
  
  +1
  
  You're welcome, happy it was helpful.

$A question about the mathematical constant e.$

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