Let's analyze and solve the limits step-by-step from the provided image.

2. $\lim_{x \to 6^+} \frac{|2 - |2 - x||}{2 - |x - 2|}$

We need to handle the absolute values carefully based on the behavior of $x$ as it approaches 6 from the right ($x \to 6^+$).

Absolute Value Analysis:

• $|2 - x|$: Since $x \to 6^+$, $x > 6$, so $|2 - x| = x - 2$.
• $|2 - (2 - x)| = |x|$: This simplifies to $x - 2$.

So, the expression becomes: $\frac{|2 - (x - 2)|}{2 - (x - 2)} = \frac{0}{x-2} \implies 0$

3. $\lim_{x \to 2} \left( \frac{1}{3 - x^2 + 5} \left( 1 - \frac{1}{1 - \sqrt{x - 1}} \right) \right)$

First, simplify the terms as $x \to 2$:

• For $\frac{1}{3 - x^2 + 5}$, substitute $x = 2$: $3 - (2)^2 + 5 = 3 - 4 + 5 = 4.$ So the fraction becomes $\frac{1}{4}$.

• Now for the second part: $1 - \frac{1}{1 - \sqrt{x - 1}}$. As $x \to 2$, $\sqrt{x - 1} = \sqrt{2 - 1} = 1$, so the expression inside becomes: $1 - \frac{1}{1 - 1} \text{ is undefined because of division by 0}.$

Thus, the limit does not exist due to a singularity at $x = 2$.

4. $\lim_{x \to 3} \frac{2 - x}{3^x + 3^{3 - x} - 28}$

Let’s first check if direct substitution works.

• Substitute $x = 3$ into the numerator: $2 - 3 = -1$.
• Now into the denominator: $3^3 + 3^{3 - 3} - 28 = 27 + 1 - 28 = 0.$

So we have the indeterminate form $\frac{-1}{0}$, which tends to $-\infty$. Therefore, the limit does not exist and goes to $-\infty$.

Would you like more details on these solutions or explanations for the methods? Here are five related questions that might help deepen your understanding:

1. What is the significance of calculating limits from the right-hand side ($x \to 6^+$)?
2. How do we deal with absolute value expressions in limits?
3. Why does the second limit lead to a singularity or undefined behavior?
4. How do we analyze limits that tend towards infinity or negative infinity?
5. What are common techniques to simplify complex limit expressions?

Tip: When evaluating limits involving absolute values, always consider the direction from which $x$ is approaching, as this affects how the absolute value expressions behave.