Approximating $e^{x}/(e^{x} - 1)$

Question

Is it correct to tell that we can approximate

\begin{equation*} \frac{e^{x}}{(e^{x} - 1)} \end{equation*}

by:

\begin{equation*} \frac{1}{x} \end{equation*}

Answer 1

Yes. But you can approximate anything by anything else. The proper way how to think about it is to consider the error. In this case, it can be proved that, for example, $$ \lim_{x\to 0} \frac{\text{First expression}}{\text{Second expression}}=1. $$ But be careful! The difference, for example, $$ \lim_{x\to 0} (\text{First expression} - \text{Second expression})=1/2 $$ is nonzero (although it is much smaller than $\infty$).

Far from $x=0$, it is definitely not a good approximation.

Answer 2

Near $x=0$, we have $$ \dfrac{e^{x}}{e^{x} - 1} = \dfrac{1}{x}+\dfrac{1}{2}+\dfrac{x}{12}-\dfrac{x^3}{720}+O(x^4) $$

Answer 3

The taylor-series $\sum_0^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$ around $a=0$ yields

$$\color{red}{\frac{1}{x}}+\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}+O(x^6)$$

However, this is only a good approximation at around $x=0$.

If you look at the graph:

you can see that the further you get away from $x=0$ the "worse" your approximation $\color{orange}{\frac{1}{x}}$becomes.