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To evaluate the limit \(\lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}}\) without using L'Hôpital's Rule, we can proceed by simplifying the expression.

### Step 1: Substitute \(x = 1\)
First, let's check what happens if we directly substitute \(x = 1\):

\[
\frac{1 - \sqrt{2 - 1^2}}{2(1) - \sqrt{2 + 2(1^2)}} = \frac{1 - \sqrt{1}}{2 - \sqrt{4}} = \frac{1 - 1}{2 - 2} = \frac{0}{0}
\]

We get the indeterminate form \(\frac{0}{0}\), so we'll need to simplify the expression further.

### Step 2: Simplify the Expression

The expression involves square roots, so we can rationalize the numerator and the denominator.

#### Rationalize the Numerator:
The numerator is \(x - \sqrt{2 - x^2}\). We rationalize by multiplying by the conjugate:

\[
\frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} \times \frac{x + \sqrt{2 - x^2}}{x + \sqrt{2 - x^2}} = \frac{(x - \sqrt{2 - x^2})(x + \sqrt{2 - x^2})}{(2x - \sqrt{2 + 2x^2})(x + \sqrt{2 - x^2})}
\]

The numerator becomes:

\[
x^2 - (2 - x^2) = 2x^2 - 2
\]

So, the expression simplifies to:

\[
\frac{2x^2 - 2}{(2x - \sqrt{2 + 2x^2})(x + \sqrt{2 - x^2})}
\]

#### Rationalize the Denominator:
Now, let's look at the denominator \(2x - \sqrt{2 + 2x^2}\) and multiply by its conjugate:

\[
\frac{2x - \sqrt{2 + 2x^2}}{2x - \sqrt{2 + 2x^2}} \times \frac{2x + \sqrt{2 + 2x^2}}{2x + \sqrt{2 + 2x^2}} = \frac{(2x - \sqrt{2 + 2x^2})(2x + \sqrt{2 + 2x^2})}{(2x + \sqrt{2 + 2x^2})(x + \sqrt{2 - x^2})}
\]

The denominator simplifies to:

\[
4x^2 - (2 + 2x^2) = 2x^2 - 2
\]

So, the limit is now:

\[
\lim_{x \to 1} \frac{2(x^2 - 1)}{2(x^2 - 1)} = \lim_{x \to 1} \frac{2(x - 1)(x + 1)}{2(x - 1)(x + 1)}
\]

### Step 3: Cancel the Common Factors
Cancel the common factors in the numerator and the denominator:

\[
\lim_{x \to 1} 1 = 1
\]

Thus, the limit is:

\[
\boxed{1}
\]

Would you like further details on any step or have any questions? 

Here are some related questions to explore:

1. How do you apply L'Hôpital's Rule to similar limits?
2. What are some alternative methods for handling indeterminate forms?
3. How does rationalizing the numerator or denominator simplify limits?
4. Can limits involving square roots always be solved by rationalization?
5. How does the concept of continuity relate to solving limits?

**Tip:** When dealing with limits, always check for common factors that can be canceled after simplifying expressions, as this often reveals the limit without needing more advanced techniques like L'Hôpital's Rule.

Learn how to evaluate the limit \(\lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}}\) without using L'Hôpital's Rule. Follow a detailed step-by-step solution that involves rationalizing the numerator and denominator to simplify the expression. Suitable for advanced undergraduate students studying limits and indeterminate forms in calculus.

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