The expression provided in the image is:

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

To solve this limit, let's proceed step by step:

Step 1: Direct Substitution

First, let's try to directly substitute $x = 1$ into the expression:

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

This results in an indeterminate form $\frac{0}{0}$, so we need to use algebraic manipulation to simplify the expression.

Step 2: Rationalization

To simplify the expression, we can rationalize both the numerator and denominator by multiplying by the conjugate of the respective terms.

Numerator:

Multiply by the conjugate of the numerator:

$\text{Conjugate of the numerator: } x + \sqrt{2 - x^2}$

So we multiply the numerator and denominator by $x + \sqrt{2 - x^2}$:

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

The numerator simplifies to:

$\left( x - \sqrt{2 - x^2} \right)\left( x + \sqrt{2 - x^2} \right) = x^2 - \left(2 - x^2\right) = 2x^2 - 2$

The denominator becomes:

$\left(2x - \sqrt{2 + 2x^2}\right)\left( x + \sqrt{2 - x^2} \right)$

Denominator:

Next, multiply the denominator by its conjugate:

$\text{Conjugate of the denominator: } 2x + \sqrt{2 + 2x^2}$

So we get:

$\frac{2x^2 - 2}{\left(2x - \sqrt{2 + 2x^2}\right)\left( x + \sqrt{2 - x^2} \right) \cdot \left(2x + \sqrt{2 + 2x^2}\right)}$

This expression is quite complicated, so we can instead try simplifying by analyzing the limits of the numerator and denominator separately or using L'Hôpital's Rule.

Step 3: Applying L'Hôpital's Rule

Since we obtained a $\frac{0}{0}$ indeterminate form, we can apply L'Hôpital's Rule. This involves taking the derivative of the numerator and denominator:

Let:

$f(x) = x - \sqrt{2 - x^2}, \quad g(x) = 2x - \sqrt{2 + 2x^2}$

Compute the derivatives:

$f'(x) = 1 - \frac{-x}{\sqrt{2 - x^2}} = 1 + \frac{x}{\sqrt{2 - x^2}}$

$g'(x) = 2 - \frac{2x}{\sqrt{2 + 2x^2}}$

So the limit becomes:

$\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{1 + \frac{x}{\sqrt{2 - x^2}}}{2 - \frac{2x}{\sqrt{2 + 2x^2}}}$

Substitute $x = 1$:

$\frac{1 + \frac{1}{\sqrt{1}}}{2 - \frac{2(1)}{2}} = \frac{1 + 1}{2 - 1} = \frac{2}{1} = 2$

Final Answer:

Thus, the limit is:

$\lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x^2}} = 2$

Would you like more details or have any further questions?

Here are 5 related questions to explore further:

1. What are some alternative methods to evaluate limits that result in indeterminate forms?
2. How does rationalizing an expression help in evaluating limits?
3. Can L'Hôpital's Rule be used for other types of indeterminate forms?
4. How do you find the derivatives of more complex functions involving square roots?
5. How does substituting different values of $x$ help in analyzing the behavior of a function near a point?

Tip: When dealing with indeterminate forms, always check if L'Hôpital's Rule applies, and ensure you differentiate correctly to avoid errors.