https://r2.erweima.ai/mathbot/image/1726421888251a8n52l5a.jpg

The expression in the image represents a limit problem:

\[
\lim_{{x \to 0}} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}
\]

### Solution:

1. **Multiply by the conjugate** to eliminate the square roots in the numerator:
   
   Multiply both the numerator and denominator by the conjugate of the numerator, \( \sqrt{1+x} + \sqrt{1-x} \):
   
   \[
   \frac{\sqrt{1+x} - \sqrt{1-x}}{x} \times \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \frac{(\sqrt{1+x})^2 - (\sqrt{1-x})^2}{x(\sqrt{1+x} + \sqrt{1-x})}
   \]

2. **Simplify the numerator**:
   
   Using the identity \( a^2 - b^2 = (a-b)(a+b) \), we get:
   
   \[
   \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{1 + x - 1 + x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}
   \]

3. **Cancel \(x\) in the numerator and denominator**:
   
   \[
   = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}
   \]

4. **Take the limit as \( x \to 0 \)**:
   
   As \( x \to 0 \), both \( \sqrt{1+x} \) and \( \sqrt{1-x} \) approach \( \sqrt{1} = 1 \). Thus:
   
   \[
   \lim_{{x \to 0}} \frac{2}{\sqrt{1+x} + \sqrt{1-x}} = \frac{2}{1 + 1} = \frac{2}{2} = 1
   \]

### Final answer:
\[
\lim_{{x \to 0}} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} = 1
\]

Would you like further details or clarification on this solution?

Here are 5 related questions to explore:
1. How do you apply L'Hopital's Rule to solve this limit?
2. Can you generalize the limit for \( \frac{\sqrt{1+ax} - \sqrt{1-bx}}{x} \)?
3. What are the conditions for using the conjugate method in limits?
4. How do square root functions behave near 0 in general?
5. How does the concept of continuity help in limit problems involving square roots?

**Tip**: Multiplying by the conjugate is a useful trick to eliminate square roots and simplify expressions, especially in limit problems.

Evaluate the limit: lim(x → 0) (sqrt(1 + x) - sqrt(1 - x)) / x

Learn how to evaluate the limit of (sqrt(1 + x) - sqrt(1 - x)) / x as x approaches 0. This problem involves using the conjugate method to simplify the expression and is suitable for students studying limits and calculus.

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