So, I am having trouble solving this limit. I have racked my brain many times to solve it. Any help is appreciated.
$$ \lim_{x \to \infty} \frac{1-e^2}{1+e^2} $$
2026-04-30 05:45:13.1777527913
Limit of $(1-e^2)/(1+e^2)$ as $x$ approches negative infinity
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If you mean $$\lim_{x\to -\infty} \frac{1-e^x}{1+e^x}$$ then the limit is equal to 1, since $$\lim_{x\to -\infty} e^x=0$$ If you did actually mean $$\lim_{x\to -\infty} \frac{1-e^2}{1+e^2}$$ then since $\frac{1-e^2}{1+e^2}$ is independent of x, the limit is simply equal to $\frac{1-e^2}{1+e^2}$.