Let's address both limits one by one.

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1. $\lim_{x \to 16} \frac{x - 16}{\sqrt{x} - 4}$

To solve this, observe that directly substituting $x = 16$ gives:

$\frac{16 - 16}{\sqrt{16} - 4} = \frac{0}{4 - 4} = \frac{0}{0},$ which is an indeterminate form. To resolve this, we'll need to simplify the expression. We can multiply both the numerator and denominator by the conjugate of the denominator.

The conjugate of $\sqrt{x} - 4$ is $\sqrt{x} + 4$. Multiply the numerator and denominator by $\sqrt{x} + 4$:

$\frac{x - 16}{\sqrt{x} - 4} \cdot \frac{\sqrt{x} + 4}{\sqrt{x} + 4} = \frac{(x - 16)(\sqrt{x} + 4)}{(\sqrt{x} - 4)(\sqrt{x} + 4)}.$

Using the difference of squares for the denominator:

$(\sqrt{x} - 4)(\sqrt{x} + 4) = x - 16.$

So, the expression becomes:

$\frac{(x - 16)(\sqrt{x} + 4)}{x - 16}.$

Now, assuming $x \neq 16$, we can cancel the $x - 16$ terms:

$\sqrt{x} + 4.$

Now, take the limit as $x \to 16$:

$\lim_{x \to 16} (\sqrt{x} + 4) = \sqrt{16} + 4 = 4 + 4 = 8.$

Thus, the limit is:

$\lim_{x \to 16} \frac{x - 16}{\sqrt{x} - 4} = 8.$

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2. $\lim_{x \to 3} \frac{x - 3}{|x - 3|}$

This limit involves an absolute value function, so we need to evaluate the left-hand and right-hand limits separately.

• For $x \to 3^+$ (approaching 3 from the right): When $x > 3$, $|x - 3| = x - 3$. The expression becomes:

  $\frac{x - 3}{x - 3} = 1.$

• For $x \to 3^-$ (approaching 3 from the left): When $x < 3$, $|x - 3| = -(x - 3)$. The expression becomes:

  $\frac{x - 3}{-(x - 3)} = -1.$

Since the left-hand and right-hand limits are not equal, the two-sided limit does not exist.

Thus, we can conclude:

$\lim_{x \to 3} \frac{x - 3}{|x - 3|} \text{ does not exist.}$

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Summary of Results:

1. $\lim_{x \to 16} \frac{x - 16}{\sqrt{x} - 4} = 8$.
2. $\lim_{x \to 3} \frac{x - 3}{|x - 3|}$ does not exist.

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Would you like more details on these problems or have any follow-up questions? Here are some related questions you might find interesting:

1. How do limits involving square roots typically behave near a point?
2. How does the absolute value affect limits at specific points?
3. What is the difference between one-sided and two-sided limits?
4. How can conjugates help simplify limits with square roots?
5. What are common techniques for evaluating indeterminate forms in limits?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ when calculating limits; they often require algebraic manipulation or special techniques like L'Hopital's Rule.