A sum of Rs. 10,000 is invested at a nominal annual interest rate of 8%, compounded continuously. (i) Derive the continuous-compounding formula $A=Pe^{rt}$ from the ordinary compound-interest formula using a limit. (ii) Using this formula, find the amount after 5 years.

Question

Accepted Answer

**Part (i) — Derivation.**

Ordinary compound interest at nominal rate $r$, compounded $n$ times a year for $t$ years:

$$A=P\left(1+\dfrac{r}{n}ight)^{nt}$$

Substitute $m=\dfrac{n}{r}$, so $n=mr$:

$$A=P\left[\left(1+\dfrac{1}{m}ight)^{m}ight]^{rt}$$

As $n	o\infty$ (continuous compounding), $m=\frac{n}{r}	o\infty$ too. By the standard limit $\displaystyle\lim_{m	o\infty}\left(1+\frac{1}{m}ight)^{m}=e$ (Section 3):

$$A=P\,e^{rt}$$

**Part (ii) — Numerical answer.**

$$A=10{,}000	imes e^{0.08	imes5}=10{,}000	imes e^{0.4}\approx10{,}000	imes1.4918\approx14{,}918$$

**Dual-check — approximate the limit numerically with a very large but finite $n$ (say $n=1{,}000$ compoundings a year) instead of using $e$ directly.** …

Q.A sum of Rs. 10,000 is invested at a nominal annual interest rate of 8%, compounded continuously.

Concept understanding — Business Applications of Limits

Q.A sum of Rs. 10,000 is invested at a nominal annual interest rate of 8%, compounded continuously.

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