What's the $\lim\limits_{x \to 1} \frac{1+e^\frac{1}{x-1}}{1-e^\frac{1}{x-1}}$

Question

$\lim\limits_{x \to 1} \frac{1+e^\frac{1}{x-1}}{1-e^\frac{1}{x-1}}$

It's always giving me $\infty$ when x approaches 1 from both sides, no matter what I do. I'm trying to solve it without l'Hopital's Rule or Taylor Series.

Answer 1

Let $t=\frac{1}{x-1} \to \infty$. Then$$\frac{1+e^t}{1-e^t} = \frac{e^{-t} +1}{e^{-t} -1} \to \frac{1}{-1} =-1$$ This is true assuming you mean $x\to 1^+$.

If $x\to 1^{-}$, then $t\to -\infty$ and the limit evaluates to $1$.

Hence, the limit doesn’t exist.