https://file.aiquickdraw.com/mathbot/file/1730919820590slocbzge.jpg

The expression you've provided is:

\[
\lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1}
\]

and it seems to imply that the limit is equal to \(\frac{1}{2}\).

Let's solve this step-by-step.

### Step-by-Step Solution

1. **Original Expression**:
   \[
   \lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1}
   \]

2. **Multiply by the Conjugate**:
   To simplify, we multiply the numerator and the denominator by the conjugate of the numerator, which is \(\sqrt{x + 1} + \sqrt{x}\):
   \[
   \lim_{{x \to \infty}} \frac{\left(\sqrt{x + 1} - \sqrt{x}\right) \cdot \left(\sqrt{x + 1} + \sqrt{x}\right)}{1 \cdot \left(\sqrt{x + 1} + \sqrt{x}\right)}
   \]

3. **Simplify the Numerator**:
   Using the difference of squares, \((a - b)(a + b) = a^2 - b^2\), we get:
   \[
   = \lim_{{x \to \infty}} \frac{(x + 1) - x}{\sqrt{x + 1} + \sqrt{x}}
   \]

   Simplifying further, the \(x\) terms cancel:
   \[
   = \lim_{{x \to \infty}} \frac{1}{\sqrt{x + 1} + \sqrt{x}}
   \]

4. **Evaluate the Denominator**:
   As \(x\) approaches infinity, both \(\sqrt{x + 1}\) and \(\sqrt{x}\) behave like \(\sqrt{x}\). So:
   \[
   \sqrt{x + 1} + \sqrt{x} \approx 2\sqrt{x}
   \]

   Thus, the expression becomes:
   \[
   \lim_{{x \to \infty}} \frac{1}{2\sqrt{x}}
   \]

5. **Take the Limit**:
   As \(x \to \infty\), \(\frac{1}{2\sqrt{x}} \to 0\).

### Conclusion
The limit is:
\[
\lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1} = 0
\]

Therefore, the correct answer is \(0\), not \(\frac{1}{2}\).

Would you like more details or have any questions?

---

Here are five related questions to deepen understanding:

1. Why do we use the conjugate method when simplifying expressions with square roots in limits?
2. How does the behavior of \(\sqrt{x}\) as \(x\) approaches infinity impact limits?
3. What other techniques, besides conjugate multiplication, can be used for evaluating limits involving radicals?
4. How would the limit change if it was \(\sqrt{x + c} - \sqrt{x}\) for some constant \(c\)?
5. What is the significance of higher-order terms when evaluating limits of functions at infinity?

**Tip**: When dealing with limits at infinity, always consider simplifying expressions using techniques like conjugate multiplication or factoring dominant terms.

Evaluate \( \lim_{{x \to \infty}} \frac{\sqrt{x+1} - \sqrt{x}}{1} \approx \frac{1}{2} \)

Discover how to evaluate the limit \( \lim_{{x \to \infty}} \frac{\sqrt{x+1} - \sqrt{x}}{1} \) with a step-by-step solution. This problem involves key concepts in limits and conjugate multiplication, providing a clear understanding of function behavior at infinity. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation