Q.Using the standard limit $\lim_{n \to \infty}\left(1+\frac1n\right)^n = e$, derive the formula for the amount under continuous compounding, and use it to find the amount to which ₹10,000 grows in 5 years at a nominal annual rate of 8%, compounded continuously. (Take $e^{0.4} \approx 1.4918$.)

Continuous compounding is what ordinary compound interest tends towards as the number of compounding periods per year is pushed to infinity, and the exponential standard limit turns that idea into an exact formula.

Deriving the formula. Under ordinary compound interest, compounded $n$ times a year at nominal annual rate $r$, a principal $P$ grows in $t$ years to $A = P\left(1+\dfrac{r}{n}\right)^{nt}$

To see what happens as compounding becomes continuous ($n \to \infty$), substitute $m = \dfrac{n}{r}$ (so $n = mr$, and as $n \to \infty$ with $r$ fixed, $m \to \infty$ too): $A = P\left(1+\dfrac1m\right)^{mrt} = P\left[\left(1+\dfrac1m\right)^{m}\right]^{rt}$

As $m \to \infty$, the bracketed term $\left(1+\dfrac1m\right)^m \to e$, by the standard limit. Therefore: $A \to P \, e^{rt}$

This is the continuous-compounding formula: $A = Pe^{rt}$

Applying it to the given data. Here $P=10{,}000$, $r=0.08$ (8% as a decimal), $t=5$ years, so $rt = 0.08 \times 5 = 0.4$ $A = 10{,}000 \times e^{0.4} \approx 10{,}000 \times 1.4918 = ₹14{,}918$ …

A common error is using $r=8$ instead of $r=0.08$ (forgetting to convert the percentage rate to a decimal) inside the exponent, which produces a wildly inflated and meaningless answer …