To find the horizontal asymptotes of the function:

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

we analyze its behavior as $x \to \pm\infty$.

Step 1: Find $\lim_{x \to \infty} f(x)$

• As $x \to \infty$, $1 - x$ behaves like $-x$.

• The exponent $\frac{x}{x - 3}$ simplifies as:

  $\frac{x}{x-3} = \frac{x}{x(1 - \frac{3}{x})} = \frac{1}{1 - \frac{3}{x}}.$

  As $x \to \infty$, $\frac{3}{x} \to 0$, so:

  $\frac{x}{x-3} \to 1.$

  This means the exponential term behaves as:

  $e^{\frac{x}{x-3}} \to e^1 = e.$

  Since $(1 - x) \approx -x$, we get:

  $f(x) \approx -x e.$

  Since $-x e \to -\infty$, there is no horizontal asymptote at $+\infty$.

Step 2: Find $\lim_{x \to -\infty} f(x)$

• As $x \to -\infty$, $1 - x \approx -x$, which grows large positively.

• The exponent:

  $\frac{x}{x-3} \to 1.$

  So, $e^{\frac{x}{x-3}} \to e$, giving:

  $f(x) \approx (-x) e.$

  Since $-x \to \infty$ as $x \to -\infty$, we get $f(x) \to \infty$, meaning there is no horizontal asymptote at $-\infty$.

Final Answer:

There are no horizontal asymptotes for this function.

---

Follow-up Questions:

1. What are the vertical asymptotes of this function?
2. How does the function behave near $x = 3$?
3. What is the derivative of this function?
4. Can we find oblique asymptotes instead of horizontal ones?
5. How does this function compare to $x e^x$?

Tip: When finding horizontal asymptotes, always examine the limits at $\pm\infty$ carefully!