Question

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

• A. 0
• B. 1
• C. 3
• D. Limit does not exist

Hint:

Recognizing the standard limit form \( \lim_{x\to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) is the fastest way to solve this type of problem. It's a fundamental limit that's worth memorizing for competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to evaluate the limit of a function as \(x\) approaches 1. Direct substitution of \(x=1\) gives the indeterminate form \(\frac{0}{0}\).

Step 2: Key Formula or Approach:
We can solve this using one of three common methods for indeterminate forms:
1.

Factorization: Factor the numerator and cancel common terms. The relevant identity is \(a^3-b^3 = (a-b)(a^2+ab+b^2)\).
2.

L'Hôpital's Rule: Differentiate the numerator and the denominator separately and then take the limit.
3.

Standard Limit Formula: Use the formula \( \lim_{x\to a} \frac{x^n - a^n}{x - a} = na^{n-1} \).

Step 3: Detailed Explanation:

Method 1: Factorization
\[ \lim_{x\to 1} \frac{x^3 - 1^3}{x - 1} = \lim_{x\to 1} \frac{(x-1)(x^2 + x \cdot 1 + 1^2)}{x - 1} \]
Cancel the \((x-1)\) term:
\[ \lim_{x\to 1} (x^2 + x + 1) \]
Now substitute \(x=1\):
\[ (1)^2 + 1 + 1 = 1 + 1 + 1 = 3 \]

Method 2: L'Hôpital's Rule
Since we have the \(\frac{0}{0}\) form, we can differentiate the numerator and denominator:
\[ \lim_{x\to 1} \frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(x - 1)} = \lim_{x\to 1} \frac{3x^2}{1} \]
Now substitute \(x=1\):
\[ \frac{3(1)^2}{1} = 3 \]

Method 3: Standard Limit Formula
The limit is in the form \( \lim_{x\to a} \frac{x^n - a^n}{x - a} \) with \(n=3\) and \(a=1\).
The result is \(na^{n-1}\):
\[ 3 \cdot (1)^{3-1} = 3 \cdot 1^2 = 3 \]

Step 4: Final Answer:
The value of the limit is 3.