I am given the function of $x(t) = \frac{x_0 e^{t}}{1+x_0(e^t-1)}$, and asked how it behaves around $0$. The solution says the $x_0(e^t-1)$ on the denominator is negligible so the function is approximated as the numerator $x(t) \simeq x_0e^{t}$ and it behaves like a exponential function around $0$.
However, can I also argue that $e^t \approx 1$ (which is used to neglect the term $x_0(e^t-1)$ on the numerator), so the whole function becomes $x_0$ around $0$? What's the problem of doing this and is there a general rule or trick that I should follow when approximating a function(evaluate the function's behavior) around a given point?
Thank you in advance for your help!
William
In the area of function approximation, you need to define what you want to preserve.
Taking your case, suppose that we want to preserve the exponential trend. So, write for example $$x(t) = \frac{x_0 e^{t}}{1+x_0(e^t-1)} = \frac{x_0+x_0 (e^{t}-1)}{1+x_0(e^t-1)}$$and define $y(t)=e^{t}-1$ to make $$x(t)=\frac{x_0+x_0 \,y(t)}{1+x_0\,y(t)}$$ and work around $y(t)=0$. By Taylor, we then have $$x(t)=x_0+x_0(1-x_0) \,y(t)+\cdots$$ that is to say $$x(t)=x_0^2+x_0(1-x_0)\, e^t$$ which is not exactly $x_0\, e^t$.
For illustration purposes, I shall use $x_0=\pi$ and say that we want to look at the behaviour for $-0.1 \leq t \leq 0.1$. As a comparison tool, let us compute the two $L^2$ norms $$\Phi_1=\int_{-0.1}^{+0.1} \Bigg[\frac{\pi e^{t}}{1+\pi(e^t-1)}-\pi e^t\Bigg]^2\,dt= 7.01\times 10^{-2}$$ $$\Phi_2=\int_{-0.1}^{+0.1} \Bigg[\frac{\pi e^{t}}{1+\pi(e^t-1)}-\left(\pi^2 +\pi(1-\pi)\,e^t\right)\Bigg]^2\,dt=2.01\times 10^{-3}$$ So $\Phi_1\sim 35\Phi_2$.
Which one do you prefer ?