Step 1: Understanding the Concept:
This limit is in the indeterminate form $\frac{0}{0}$ because substituting $x=1$ results in $\frac{1^3-1}{1-1} = \frac{0}{0}$. We can solve this either by algebraic factorization or by using L'Hôpital's Rule.
Step 2: Key Formula or Approach:
Method 1: Factorization
Use the algebraic identity for the difference of cubes: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$.
Method 2: L'Hôpital's Rule
If $\lim_{x\to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$.
Step 3: Detailed Explanation:
Using Factorization:
Let's factor the numerator $x^3-1$:
\[ x^3 - 1^3 = (x-1)(x^2 + x \cdot 1 + 1^2) = (x-1)(x^2+x+1) \]
Now substitute this back into the limit expression:
\[ \lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{x-1} \]
For $x \neq 1$, we can cancel the $(x-1)$ terms:
\[ \lim_{x\to 1} (x^2+x+1) \]
Now, substitute $x=1$:
\[ = (1)^2 + 1 + 1 = 1+1+1 = 3 \]
Using L'Hôpital's Rule:
The limit is of the form $\frac{0}{0}$. We can differentiate the numerator and the denominator separately.
\[ f(x) = x^3 - 1 \implies f'(x) = 3x^2 \]
\[ g(x) = x - 1 \implies g'(x) = 1 \]
The new limit is:
\[ \lim_{x\to 1} \frac{3x^2}{1} \]
Substitute $x=1$:
\[ = \frac{3(1)^2}{1} = 3 \]
Step 4: Final Answer:
Both methods yield a result of 3. Therefore, option (C) is correct.