To find the limit of the function

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

as $x \to -\infty$:

1. Behavior of the numerator $e^x$:
   As $x \to -\infty$, $e^x \to 0^+$ (exponentially fast decay).

2. Behavior of the denominator $x - 1$:
   As $x \to -\infty$, $x - 1 \to -\infty$.

3. Combining the behavior:
   Since the numerator approaches $0$ and the denominator approaches $-\infty$, the fraction approaches

   $\frac{0}{-\infty} = 0.$

Thus, the limit is

$\lim_{x \to -\infty} \frac{e^x}{x - 1} = 0.$

Let me know if you'd like further clarifications!

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Related Questions:

1. What is the limit of $\frac{e^x}{x}$ as $x \to -\infty$?
2. How does $\frac{e^x}{x-1}$ behave as $x \to +\infty$?
3. Can $e^x$ ever approach $1$ or another constant as $x \to -\infty$?
4. What happens to the function $\frac{e^x}{x - a}$ for different constants $a$?
5. How do you find horizontal asymptotes for rational functions involving $e^x$?

Tip:

For functions involving $e^x$, always analyze the growth or decay of $e^x$ relative to other terms in the function.