Let's go through the solutions for each limit step-by-step:

Problem 1:

$\lim_{{x \to 1}} \frac{\sqrt{x+3} - 2}{x - 1}$

1. Direct Substitution: Substitute $x = 1$. $\frac{\sqrt{1+3} - 2}{1 - 1} = \frac{2 - 2}{0} = \frac{0}{0}$ This is an indeterminate form, so we need to manipulate the expression.

2. Multiply by the Conjugate: Multiply the numerator and the denominator by the conjugate of the numerator, $\sqrt{x+3} + 2$. $\frac{\sqrt{x+3} - 2}{x - 1} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2} = \frac{(x+3) - 4}{(x - 1)(\sqrt{x+3} + 2)}$

3. Simplify: The numerator becomes $x - 1$. $= \frac{x - 1}{(x - 1)(\sqrt{x+3} + 2)}$

4. Cancel $x - 1$ from the numerator and denominator. $= \frac{1}{\sqrt{x+3} + 2}$

5. Substitute $x = 1$. $= \frac{1}{\sqrt{1+3} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

So, the answer for the first limit is: $\lim_{{x \to 1}} \frac{\sqrt{x+3} - 2}{x - 1} = \frac{1}{4}$

Problem 2:

$\lim_{{x \to 1}} \frac{x^3 + 1}{x^2 - 1}$

1. Direct Substitution: Substitute $x = 1$. $\frac{1^3 + 1}{1^2 - 1} = \frac{2}{0}$ This gives an undefined form, so we need to simplify.

2. Factor the Expressions:

   • For the numerator, $x^3 + 1$ is a sum of cubes: $x^3 + 1 = (x + 1)(x^2 - x + 1)$.
   • For the denominator, $x^2 - 1$ is a difference of squares: $x^2 - 1 = (x + 1)(x - 1)$.

   So we can rewrite the expression as: $\frac{(x + 1)(x^2 - x + 1)}{(x + 1)(x - 1)}$

3. Cancel $x + 1$ from the numerator and denominator. $= \frac{x^2 - x + 1}{x - 1}$

4. Substitute $x = 1$.

   • We still have a $\frac{0}{0}$ form.