The problem is asking to find the limit of the following expression as $x \to \infty$:

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

Step-by-Step Solution:

1. Factor the expression inside the square root: $\sqrt{x^2 + 1}$ To simplify this, factor $x^2$ from the terms inside the square root: $\sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = |x| \sqrt{1 + \frac{1}{x^2}}$

2. Simplify as $x \to \infty$: As $x \to \infty$, the term $\frac{1}{x^2}$ approaches 0: $\sqrt{1 + \frac{1}{x^2}} \to \sqrt{1} = 1$ Therefore, for large $x$, $\sqrt{x^2 + 1} \approx |x| \cdot 1 = |x|$

   Since $x$ is approaching $+\infty$, $|x| = x$.

3. Conclusion: Therefore, the limit is: $\lim_{x \to \infty} \sqrt{x^2 + 1} = x = \infty$

So, the limit does not exist (DNE) in the usual sense, because the expression grows without bound as $x$ approaches infinity.

Would you like more details or have any questions on this? Here are some related questions to deepen your understanding:

1. What would the limit be if the expression was $\lim_{x \to \infty} \frac{\sqrt{x^2 + 1}}{x}$?
2. How would the answer change if $x \to -\infty$ instead of $+\infty$?
3. Why does the square root $\sqrt{x^2 + 1}$ grow similarly to $x$ for large values of $x$?
4. How does factoring help simplify limits involving square roots?
5. Can you give an example where the limit of a similar function exists?

Tip: When dealing with limits at infinity, it's often useful to factor out the highest power of $x$ from inside functions like square roots to simplify the expression.