The given function is:

$f(x) = \frac{2x}{\sqrt{x^2 + x + 1}}$

To find the horizontal asymptote of this function, we need to analyze its behavior as $x \to \infty$ and $x \to -\infty$.

Step 1: Analyze $f(x)$ as $x \to \infty$

When $x$ becomes very large, the $x^2$ term dominates in both the numerator and the denominator. Therefore, we approximate the function for large $x$ by ignoring the lower-order terms in the denominator: $f(x) \approx \frac{2x}{\sqrt{x^2}} = \frac{2x}{|x|}.$

For $x \to \infty$, we have $|x| = x$, so: $f(x) \approx \frac{2x}{x} = 2.$ Thus, as $x \to \infty$, the horizontal asymptote is $y = 2$.

Step 2: Analyze $f(x)$ as $x \to -\infty$

Similarly, for $x \to -\infty$, we again approximate the function: $f(x) \approx \frac{2x}{\sqrt{x^2}} = \frac{2x}{|x|}.$

For $x \to -\infty$, $|x| = -x$, so: $f(x) \approx \frac{2x}{-x} = -2.$ Thus, as $x \to -\infty$, the horizontal asymptote is $y = -2$.

Conclusion

The function has two horizontal asymptotes:

• As $x \to \infty$, $y = 2$.
• As $x \to -\infty$, $y = -2$.

Would you like further details on the process or have any other questions?

Here are some related questions:

1. What is the significance of horizontal asymptotes in a function's graph?
2. How would the function's asymptotes change if the numerator were different?
3. What are vertical asymptotes, and how are they determined?
4. How do you find slant asymptotes if they exist?
5. What happens to the behavior of $f(x)$ for values of $x$ near 0?

Tip: When finding horizontal asymptotes, focus on the behavior of the function for very large positive and negative values of $x$, as this reveals the long-term trend.