To find the horizontal asymptotes of the given function ( f(x) = x + \sqrt{xThe given function is:

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

Steps to Find Horizontal Asymptotes:

We need to examine the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$.

1. As $x \to \infty$:

For large positive values of $x$, we can factor $x^2$ out of the square root:

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

Using the property $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$, we simplify:

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

For large $x$, $\sqrt{1 - \frac{1}{x^2}} \approx 1 - \frac{1}{2x^2}$ (using the binomial expansion for small $\frac{1}{x^2}$):

$f(x) \approx x + x\left(1 - \frac{1}{2x^2}\right) = x + x - \frac{1}{2x}.$

So, as $x \to \infty$, the dominant term is $2x$. There is no horizontal asymptote in this direction because $f(x)$ grows without bound.

2. As $x \to -\infty$:

For large negative values of $x$, the same steps apply because $x^2 - 1$ is still positive, and the square root simplifies similarly:

$f(x) = x + \sqrt{x^2 - 1} \approx x + x = 2x.$

Again, as $x \to -\infty$, $f(x) \to 2x$, which means there is no horizontal asymptote in this direction either.

Conclusion:

The function $f(x)$ does not have horizontal asymptotes because it grows linearly as $x \to \pm\infty$.

Do you want me to clarify this further or explore related concepts?

Here are some related questions for further exploration:

1. What are vertical asymptotes, and does this function have any?
2. How do you find slant (oblique) asymptotes for functions like this?
3. What is the domain of the given function $f(x)$?
4. Can $f(x)$ ever intersect its asymptotic behavior (if extended)?
5. How does the derivative of $f(x)$ behave as $x \to \infty$?

Tip: When analyzing the behavior of functions at infinity, always focus on dominant terms!