The question is asking for the horizontal asymptotes of the function $f(x) = 1 - e^{-\frac{1}{x}}$.

Step-by-Step Solution:

1. Understand horizontal asymptotes: Horizontal asymptotes describe the behavior of a function as $x \to \infty$ or $x \to -\infty$. To find them, calculate the limits of $f(x)$ as $x \to \infty$ and $x \to -\infty$.

2. Behavior as $x \to \infty$:

   • As $x \to \infty$, $\frac{1}{x} \to 0^+$. Hence: $e^{-\frac{1}{x}} \to e^0 = 1.$ Substituting this into $f(x)$: $f(x) = 1 - e^{-\frac{1}{x}} \to 1 - 1 = 0.$ Therefore, as $x \to \infty$, the horizontal asymptote is $y = 0$.

3. Behavior as $x \to -\infty$:

   • As $x \to -\infty$, $\frac{1}{x} \to 0^-$. Hence: $e^{-\frac{1}{x}} \to e^0 = 1.$ Substituting this into $f(x)$: $f(x) = 1 - e^{-\frac{1}{x}} \to 1 - 1 = 0.$ Therefore, as $x \to -\infty$, the horizontal asymptote is also $y = 0$.

4. Conclusion: The function has a single horizontal asymptote at: $y = 0.$

Would you like a deeper explanation or have additional questions about the process?

Related Questions:

1. What are vertical asymptotes, and does this function have any?
2. How does the exponential function $e^{-\frac{1}{x}}$ behave for large positive and negative $x$?
3. Can a function have more than one horizontal asymptote?
4. What is the significance of the derivative of $f(x)$ in analyzing its behavior near asymptotes?
5. How does the concept of limits help define asymptotes?

Tip:

When finding horizontal asymptotes, focus on the behavior of the function as $x \to \infty$ and $x \to -\infty$; exponential functions often simplify significantly in such cases.