I did quite a bit of research before getting stuck here. So I have a function: $\frac{e^x}{cos(x)}$
A vertical asymptote occurs every $n\pi+\frac{\pi}{2}$ I believe (as the denominator approaches 0), but we also have $\lim_{x\to -\infty} e^x = 0$ and since it's essentially $\frac{1}{e^{|x|}\cos(x)}$ so as $cos(x)$ approaches zero in its cycle, $f(x)$ will approach infinity.
Is $\lim_{x\to -\infty} \frac{e^x}{\cos(x)} = 0$ still a horizontal asymtote despite this?
No, because you still have vertical asymptotes every $\frac{\pi}{2}$ as $x\to-\infty$. To have a horizontal asymptote you would need to get closer and closer to $0$ and stay close.