The image you provided contains the following mathematical limit problem:

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

The limit evaluates to an indeterminate form $\frac{0}{0}$, as shown in the image. To solve this, we will apply some techniques to resolve the indeterminate form. Let's proceed by using the method of multiplying by the conjugate of the numerator to simplify the expression.

The conjugate of $\sqrt{2 + |x - 1|} - \sqrt{2}$ is $\sqrt{2 + |x - 1|} + \sqrt{2}$. We will multiply both the numerator and denominator by this conjugate to simplify the square root terms.

The expression becomes:

$\lim_{{x \to 1}} \frac{\left( \sqrt{2 + |x - 1|} - \sqrt{2} \right) \left( \sqrt{2 + |x - 1|} + \sqrt{2} \right)}{\left(1 - x^2 \right)\left( \sqrt{2 + |x - 1|} + \sqrt{2} \right)}$

This simplifies the numerator:

$\left( \sqrt{2 + |x - 1|} \right)^2 - \left( \sqrt{2} \right)^2 = (2 + |x - 1|) - 2 = |x - 1|$

So the expression simplifies to:

$\lim_{{x \to 1}} \frac{|x - 1|}{(1 - x^2) \left( \sqrt{2 + |x - 1|} + \sqrt{2} \right)}$

Now, we rewrite $1 - x^2$ as $(1 - x)(1 + x)$, yielding:

$\lim_{{x \to 1}} \frac{|x - 1|}{(1 - x)(1 + x) \left( \sqrt{2 + |x - 1|} + \sqrt{2} \right)}$

Since $|x - 1| = -(x - 1)$ for $x \to 1^-$ and $|x - 1| = (x - 1)$ for $x \to 1^+$, the absolute value cancels out, and we are left with:

$\lim_{{x \to 1}} \frac{-1}{(1 + 1) \left( \sqrt{2 + 0} + \sqrt{2} \right)} = \frac{-1}{2 \times 2\sqrt{2}} = \frac{-1}{4\sqrt{2}}$

Thus, the value of the limit is:

$\boxed{\frac{-1}{4\sqrt{2}}}$

Would you like more details on any part of this solution, or have any questions?

Here are some related questions for further exploration:

1. How do we apply the method of conjugates in other limit problems?
2. What are other indeterminate forms besides $\frac{0}{0}$?
3. Can L'Hopital's Rule be applied to this problem? Why or why not?
4. How does the behavior of absolute value functions affect the limit evaluation?
5. What is the significance of simplifying complex fractions in limit problems?

Tip: When working with limits involving square roots, multiplying by the conjugate is a helpful technique to eliminate radicals and simplify the expression.