limit of [{x-√(2-x^2)} / {2x-√(2+2x^)}]

To evaluate the limit

\[
\lim_{{x \to 1}} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x}}
\]

we’ll start by directly substituting \( x = 1 \) into the expression to see if it results in an indeterminate form:

1. **Substitute \( x = 1 \):**

   \[
   \frac{1 - \sqrt{2 - 1^2}}{2 \cdot 1 - \sqrt{2 + 2 \cdot 1}} = \frac{1 - \sqrt{2 - 1}}{2 - \sqrt{2 + 2}}
   \]

   Simplifying inside the square roots:

   \[
   = \frac{1 - \sqrt{1}}{2 - \sqrt{4}} = \frac{1 - 1}{2 - 2} = \frac{0}{0}
   \]

   This results in the indeterminate form \( \frac{0}{0} \), so we need to manipulate the expression further to evaluate the limit.

2. **Rationalize the numerator and denominator:**

   To resolve the indeterminate form, we can multiply both the numerator and the denominator by the conjugate expressions of each part.

   **Rationalize the numerator:** Multiply the numerator and the denominator by the conjugate of \( x - \sqrt{2 - x^2} \), which is \( x + \sqrt{2 - x^2} \):

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

   This simplifies the numerator to:

   \[
   (x)^2 - ( \sqrt{2 - x^2} )^2 = x^2 - (2 - x^2) = x^2 - 2 + x^2 = 2x^2 - 2 = 2(x^2 - 1)
   \]

   Therefore, the expression becomes:

   \[
   \frac{2(x^2 - 1)}{ \left( 2x - \sqrt{2 + 2x} \right) \left( x + \sqrt{2 - x^2} \right) }
   \]

   Notice that \( x^2 - 1 = (x - 1)(x + 1) \), so we can rewrite it as:

   \[
   \frac{2(x - 1)(x + 1)}{ \left( 2x - \sqrt{2 + 2x} \right) \left( x + \sqrt{2 - x^2} \right) }
   \]

3. **Simplify the expression and apply the limit:**

   We can now proceed by further rationalizing the denominator in a similar manner if necessary.

limit of [(x - √(2 - x^2)) / (2x - √(2 + 2x))]

Explore the step-by-step solution for finding the limit of [(x - √(2 - x^2)) / (2x - √(2 + 2x))] as x approaches 1. This calculus problem involves handling indeterminate forms and rationalizing expressions, suitable for Grade 12 or college-level calculus students.

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