Proof of continuous force of interest with infinitely compounded interest rate. For actuaries, delta and i upper infinity

Question

Specifically, we know the following: $$(1+\frac{i^{(2)}}{2})^2 =(1+\frac{i^{(4)}}{4})^4=(1+\frac{i^{(12)}}{12})^{12}= 1+i, $$ Where $i^{(2)}$, $i^{(4)}$, and $i^{(12)}$ are the interest rates compunded semiannually, quarterly, and monthly, respectively, and i is the interest rate compunded annually. However, I noticed in the Standard Ultimate Life Table provided by the Society of Actuaries that $i^{(\infty)}$ = $\delta$, or more specifically: $$\lim\limits_{m \to \infty}(1+\frac{i^{(m)}}{m})^m=e^\delta=1+i$$ This makes perfect sense intuitively and I found a post about the more common notation $A=Pe^{rt}$ where $\lim\limits_{n \to \infty}(1+\frac{1}{n})^n=e$ is used in the proof to show the relationship with $A=P(1+\frac{r}{n})^{nt}$, which also makes sense, but I was wondering if someone could use the relationship bewteen $i^{(m)}$ and i to prove the following: $$\lim\limits_{m \to \infty}[(1+i)^{\frac{1}{m}}-1]m=\delta$$Where $\delta=\ln(1+i)$. Thanks!

Answer 1

Note $$\left((1+i)^{1/m} - 1\right)m = \log(1+i) \cdot \frac{\exp\left(\frac{1}{m} \log (1+i)\right) - e^0}{\frac{1}{m} \log(1+i) - 0},$$ so that with the substitution $h = \frac{1}{m} \log(1+i)$, we obtain $$\lim_{m \to \infty} \left((1+i)^{1/m} - 1\right)m = \log(1+i) \lim_{h \to 0} \frac{e^h - e^0}{h - 0} = f'(0) \log(1+i) = \log(1+i),$$ where $f(x) = e^x$ and we have applied the definition of derivative $$f'(a) = \lim_{h \to a} \frac{f(h) - f(a)}{h - a}.$$