The mathematical expression provided in the image is:

$\lim_{x \to 2} \frac{\sqrt{x-2 + x^2 -3x + 2}}{\sqrt{x^2 - 4}}$

To evaluate this limit, we can simplify the expressions in the numerator and the denominator:

1. Numerator: $x - 2 + x^2 - 3x + 2$ simplifies to $x^2 - 2x + 2$.

2. Denominator: $x^2 - 4$ can be factored using the difference of squares formula: $(x-2)(x+2)$.

Substituting back into the limit expression:

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

Evaluating directly by substituting $x = 2$:

• In the numerator: $\sqrt{2^2 - 2 \cdot 2 + 2} = \sqrt{2}$.
• In the denominator: $\sqrt{(2-2)(2+2)} = \sqrt{0 \cdot 4} = 0$.

This forms a $\frac{0}{0}$ indeterminate form, suggesting we need to apply L'Hôpital's rule or simplify further.

Applying L'Hôpital's Rule:

[ \lim_{x \to 2} \frac{\frac{d}{dx}[\sqrt{x^2 - 2x + 2}]}{\frac{d}{dx}[\sqrt{(x-2)(x+2)}]} ]

Let's compute the derivatives of the numerator and denominator:

• Numerator Derivative: Use the chain rule: $\frac{d}{dx}[\sqrt{x^2 - 2x + 2}] = \frac{1}{2\sqrt{x^2 - 2x + 2}} \cdot 2(x - 1) = \frac{x-1}{\sqrt{x^2 - 2x + 2}}$

• Denominator Derivative: Again use the chain rule: $\frac{d}{dx}[\sqrt{(x-2)(x+2)}] = \frac{1}{2\sqrt{(x-2)(x+2)}} \cdot (2x) = \frac{x}{\sqrt{(x-2)(x+2)}}$

Substituting these back into the limit gives:

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

Simplifying further:

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

We substitute $x = 2$ directly:

$\frac{2-1}{2} \cdot \frac{\sqrt{0}}{\sqrt{2}} = 0$

Thus, the limit is $0$.

Would you like to delve deeper into any part of this process, or do you have any other questions on different topics?

Further Exploration Questions:

1. How does L'Hôpital's Rule apply when the limit still results in an indeterminate form after the first application?
2. What are other techniques to handle limits involving square roots?
3. How do we factor quadratic expressions like $x^2 - 2x + 2$?
4. Can we discuss the geometrical interpretation of the limit problem?
5. What is the significance of indeterminate forms in calculus?

Tip: When dealing with complex limits, always consider simplifying expressions and checking for indeterminate forms before applying advanced techniques like L'Hôpital's Rule.