We are asked to find the limit of the following expression as $x$ approaches infinity:

$\lim_{{x \to \infty}} \frac{x}{\sqrt{x^2 + 1}}$

Step-by-Step Solution:

1. Factor $x^2$ out of the square root:

   Rewrite the expression inside the square root:

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

   So the expression becomes:

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

2. Evaluate the limit as $x \to \infty$:

   As $x \to \infty$, the term $\frac{1}{x^2} \to 0$. Therefore, the expression simplifies to:

   $\lim_{{x \to \infty}} \frac{1}{\sqrt{1 + \frac{1}{x^2}}} = \frac{1}{\sqrt{1 + 0}} = 1$

Thus, the limit is:

$\boxed{1}$

Would you like more details on this solution, or do you have any further questions?

Here are 5 related questions to explore:

1. What is the limit of $\frac{x^2}{\sqrt{x^4 + 1}}$ as $x \to \infty$?
2. How do you handle limits involving radicals and infinity in general?
3. What is the limit of $\frac{x}{\sqrt{x^2 + x}}$ as $x \to \infty$?
4. Can L'Hopital's Rule be applied to find the limit here? If so, how?
5. How does the behavior of functions change as $x \to -\infty$?

Tip: When finding limits involving square roots at infinity, factor out the highest power of $x$ to simplify the expression.