No idea if this is the right stackexchange for this, feel free to point me elsewhere!

I'm teaching business calculus and one of the problems the students have is to figure out the present value of a continuous income stream $\$$$A$ per year earning $r$ interest rate. The integral comes to $$P.V=A\int_0^te^{-rz}dz=\frac A {-r}(e^{-rt}-1)$$ I've noticed as a heuristic that for relatively short values of t ($t<10$) and small values of r ($r<.04$) (the parameters the learning software uses), the present value comes to be just a bit less than $A\cdot t$. (usually within about 10 percent) Does anyone know an economics rational why this is the case, and would it hold for larger values of $r$ or $t$?


1 Answer 1



Well, when $r$ is zero, you are just getting value $A$ per time unit, so total present value will be $A \cdot t$.

The approximation

You are asking about the approximation $$ \frac A {-r}(e^{-rt}-1) \approx A \cdot t. $$ This is equivalent to $$ \frac 1{rt}(1-e^{-rt}) \approx 1, $$ which is true for small $rt$ values. You can see this by taking the limit of the left hand side and using L'Hospital's rule. $$ \lim_{rt \to 0} \frac{\frac{\text{d} (1 - e^{-rt})}{\text{d} rt}}{\frac{\text{d} rt}{\text{d} rt}} = \lim_{rt \to 0} \frac{e^{-rt}}{1} = 1. $$

The approximation in another form

The approximation $$ 1-e^{-h} \approx h $$ seems to be a variant of the more well known approximation $$ \ln(1+h) \approx h $$ for small values of $h$.

How good is this approximation

To get a sense of the error term for large values using your approximation, you can use Taylor-approximation for the function $$ f(x) = \frac{1-e^{-x}}{x} $$ starting from some very small $x$ value.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.