Let's evaluate the following limit:

$\lim_{x \to \infty} \left(\frac{x}{\sqrt{x^2 + 1}}\right)$

Step 1: Factor the denominator

Notice that the expression inside the square root in the denominator is $x^2 + 1$. To simplify the limit as $x \to \infty$, let's factor $x^2$ from inside the square root:

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

For large $x$, we know that $|x| = x$ (since $x$ is positive as $x \to \infty$). Therefore, we can rewrite the expression as:

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

Step 2: Take the limit

Now, as $x \to \infty$, the term $\frac{1}{x^2}$ becomes very small, and we have:

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

Final Answer:

$\lim_{x \to \infty} \left(\frac{x}{\sqrt{x^2 + 1}}\right) = 1$

Would you like more details on this process or have any questions? Here are some related questions to expand on this topic:

1. What happens to the limit of $\frac{x}{\sqrt{x^2 + c}}$ for different constants $c$?
2. How do you evaluate limits that involve square roots in the denominator?
3. What happens to the limit $\lim_{x \to \infty} \left( \frac{1}{\sqrt{x^2 + 1}} \right)$?
4. How can this type of limit be generalized for functions of the form $\frac{p(x)}{q(x)}$ where both $p(x)$ and $q(x)$ are polynomials?
5. What is the meaning of the square root approximation $\sqrt{1 + \epsilon} \approx 1 + \frac{\epsilon}{2}$ for small $\epsilon$?

Tip: Whenever you have a square root involving a large variable, like $x$, try factoring out the highest power of $x$ inside the square root. This often simplifies the expression, especially in limits as $x \to \infty$.