To solve the given limit:

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

we notice that substituting $x = 1$ directly leads to an indeterminate form $\frac{0}{0}$. This requires further simplification.

Step 1: Simplify the numerator

The numerator is $\sqrt{x + 3} - 2$. To simplify, we multiply and divide by the conjugate:

$\sqrt{x + 3} - 2 = \frac{(\sqrt{x + 3} - 2)(\sqrt{x + 3} + 2)}{\sqrt{x + 3} + 2}.$

This simplifies to:

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

Step 2: Factorize the denominator

The denominator is $x^3 - 1$, which is a difference of cubes:

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

Step 3: Simplify the expression

Substituting the simplified numerator and factorized denominator, the limit becomes:

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

Cancel $x - 1$ from the numerator and denominator (valid since $x \neq 1$):

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

Step 4: Substitute $x = 1$

Now, substitute $x = 1$ into the remaining expression:

$\frac{1}{(\sqrt{1 + 3} + 2)((1)^2 + 1 + 1)} = \frac{1}{(2 + 2)(1 + 1 + 1)} = \frac{1}{4 \cdot 3} = \frac{1}{12}.$

Final Answer:

$\lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x^3 - 1} = \frac{1}{12}.$

Let me know if you'd like a detailed breakdown of any step or further questions!

Related Questions:

1. How do you use the conjugate method for simplifying limits?
2. Can all indeterminate forms be solved using simplification?
3. What are other techniques to handle limits involving radicals?
4. How does the difference of cubes formula help in calculus problems?
5. How do you interpret the result of a limit geometrically?

Tip: Always check for factorization opportunities and conjugate tricks when handling indeterminate forms in limits!