Let's go through a few of these limit problems one by one.

Problem 19:

$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$ To solve this limit, notice that directly substituting $x = 1$ leads to a $\frac{0}{0}$ indeterminate form. So, we can rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator:

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

Now, as $x \to 1$, we get:

$\frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Thus, the limit is: $\boxed{\frac{1}{2}}$

Problem 20:

$\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$ Again, substituting $x = 1$ gives a $\frac{0}{0}$ indeterminate form. We can factor both the numerator and denominator:

$x^3 - 1 = (x - 1)(x^2 + x + 1)$ $x^2 - 1 = (x - 1)(x + 1)$

Cancel out the common factor $x - 1$:

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

Now, substituting $x = 1$:

$\frac{1^2 + 1 + 1}{1 + 1} = \frac{3}{2}$

Thus, the limit is: $\boxed{\frac{3}{2}}$

Would you like me to continue solving the remaining problems? Let me know if you'd like any specific details or more problems explained!

Here are some follow-up questions:

1. How can limits help in understanding the behavior of functions near points of discontinuity?
2. What are common techniques to resolve indeterminate forms like $\frac{0}{0}$?
3. How do conjugates work in rationalizing expressions?
4. Why is factoring an effective method in solving limit problems?
5. What are other forms of indeterminate expressions besides $\frac{0}{0}$?

Tip: Always check for common factors in the numerator and denominator when dealing with rational expressions before applying limits.