Vertical asymptotes of $f(x)=\frac{e^x-3}{e^{2x}+e^x-1}$

Question

I'm trying to study the graph of the function

$$f(x)=\frac{e^x-3}{e^{2x}+e^x-1}$$

and I'm having some trouble finding the vertical asymptotes of it. So before skipping to the asymptotes progress I'll just iterate over the discontinuity points I've found. We have that the domain of the function is

$$\mathbb{R}-\{e^{2x}+e^x-1 = 0\}$$

Where $e^{2x}+e^x-1 = 0$ solves for

$$e^x=\frac{-1\pm\sqrt{5}}{2}$$

Where $\frac{-1-\sqrt{5}}{2}$ isn't a possible solution since logarithms of negative numbers do not exist. This brings us to state that the domain of $f(x)$ is

$$\mathbb{R}-\{\log{\frac{\sqrt{5}-1}{2}}\}$$

where $\log{\frac{\sqrt{5}-1}{2}}\approx-\frac12$.

At this point to find the vertical asymptotes we have to find the limit:

$$\lim_{x\rightarrow\log{\frac{\sqrt{5}-1}{2}}}\frac{e^x-3}{e^{2x}+e^x-1}$$

It plays out nicely (I guess) since $e^{\log{x}}=x$ so the limit becomes

$$\frac{-5+\sqrt5}{3+\sqrt5}$$

This however doesn't make sense since plotting the graph on geogebra says clearly that on $\log{\frac{\sqrt{5}-1}{2}}$ there is clearly a vertical asymptote where for $x\rightarrow\log{\frac{\sqrt{5}-1}{2}}^-$ $y\rightarrow+\infty$ while for $x\rightarrow\log{\frac{\sqrt{5}-1}{2}}^+$ $y\rightarrow-\infty$

It's weird because the horizontal ones are correct, the discontinuity point matches with the one in geogebra and the sign function study plays out correctly too. I'm just having an issue with proving that there is a vertical asymptote. Where did I go wrong?

Here's the geogebra graph

Answer 1

Here is why you made a mistake:

the roots of $X^2 + X -1$ are indeed $r_{\pm} = -\frac{1 \pm \sqrt{5}}{2}$, so it is natural to study near $\ln(r_-)$.

If you compose with $f$ with $\ln(x)$, you will get $f\circ \ln (x) = \frac{x-3}{x^2 + x -1} = \frac{x-3}{(x-r_+)(x-r_-)}$. So your computations are wrong. This diverges clearly at $r_- $, $ \infty$ on the left and $-\infty$ on the right.

Since $\ln$ is continuous at $r_-$ and increasing, the limit of you function at $r_-$ will be the same as the limit of $f$ at $\ln(r_-)$. So the behavior of $f$ is the same.