Question

The value of \(\lim_{x \to 1} \frac{x^3-1}{x-1}\) is

• A. \(\infty\)
• B. 0
• C. 1
• D. 3

Hint:

For limits of rational functions that result in \(\frac{0}{0}\), factoring is often the quickest method if you can spot the algebraic identity. L'Hôpital's rule is a powerful alternative, especially for more complex functions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given limit is of the indeterminate form \(\frac{0}{0}\) because substituting \(x=1\) into the numerator and denominator gives \(\frac{1^3-1}{1-1} = \frac{0}{0}\). We can solve this by either factoring the numerator or using L'Hôpital's Rule.

Step 2: Key Formula or Approach:
Method 1: Factoring
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 Factoring:
We factor the numerator \(x^3-1\) as \(x^3 - 1^3\): \[ x^3 - 1 = (x-1)(x^2 + x . 1 + 1^2) = (x-1)(x^2+x+1) \] 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, we can substitute \(x=1\) into the simplified expression: \[ 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \] Using L'Hôpital's Rule:
Let \(f(x) = x^3 - 1\) and \(g(x) = x - 1\).
The derivatives are \(f'(x) = 3x^2\) and \(g'(x) = 1\).
Applying the rule: \[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{3x^2}{1} \] Substituting \(x=1\): \[ \frac{3(1)^2}{1} = 3 \]
Step 4: Final Answer:
Both methods yield the same result. The value of the limit is 3.