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Exercise 7.5 · Q12

Q.If an initial amount A0A_0 of money is invested at an interest rate rr compounded nn times a year, the value of the investment after tt years is A=A0left(1+dfracrnright)ntA=A_0\\left(1+\\dfrac{r}{n}\\right)^{nt}. If the interest is compounded continuously, (that is as ntoinftyn\\to\\infty), show that the amount after tt years is A=A0ertA=A_0e^{rt}.

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Take logarithms of the compounding formula, substitute u=r/n→0u=r/n\to0 as n→∞n\to\infty, evaluate the resulting standard ln⁡(1+u)u→1\tfrac{\ln(1+u)}{u}\to1 limit via l'Hôpital, then exponentiate back.

Step 1. Set up the limit.

A=A0(1+rn)ntA=A_0\left(1+\dfrac{r}{n}\right)^{nt}. Let L=lim⁡n→∞(1+rn)ntL=\displaystyle\lim_{n\to\infty}\left(1+\dfrac{r}{n}\right)^{nt}, so A→A0LA\to A_0L as n→∞n\to\infty.

Step 2. Take logarithms.

log⁡L=lim⁡n→∞ntlog⁡(1+rn).\log L=\lim_{n\to\infty}nt\log\left(1+\frac{r}{n}\right).

Step 3. Substitute u=rnu=\dfrac{r}{n} (so n=run=\dfrac{r}{u}, and u→0u\to0 as n→∞n\to\infty).

ntlog⁡(1+rn)=ru t log⁡(1+u)=rt⋅log⁡(1+u)u.nt\log\left(1+\frac{r}{n}\right)=\frac{r}{u}\,t\,\log(1+u)=rt\cdot\frac{\log(1+u)}{u}.

Step 4. Evaluate lim⁡u→0log⁡(1+u)u\displaystyle\lim_{u\to0}\frac{\log(1+u)}{u} (a 00\tfrac00 form) via l'Hôpital. …

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