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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.

Find the limit of lim (x→∞) √(x² + 1), if it exists.

Find the limit of the square root function lim (x→∞) √(x² + 1). This problem demonstrates concepts of limits and asymptotic behavior. Learn how factoring helps simplify the expression and solve for infinity.

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