Question

If \[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}, \text{ then find } k \]

• A. \( \frac{8}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{2}{3} \)
• D. 1

Hint:

To solve limit problems, factor the expressions and cancel common factors wherever possible before substituting the value.

Answer 1

The Correct Option is A

Step 1: Consider the left-hand side of the equation:
\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \] Factor \( x^4 - 1 \) as \( (x^2 - 1)(x^2 + 1) \), and further factor \( x^2 - 1 \) as \( (x - 1)(x + 1) \). Thus, we get: \[ \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1} \] Cancel the common factor \( x - 1 \): \[ (x + 1)(x^2 + 1) \] Now substitute \( x = 1 \): \[ (1 + 1)(1^2 + 1) = 2 \times 2 = 4 \] Thus, the left-hand side gives 4. Step 2: Now consider the right-hand side of the equation: \[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \] Factor \( x^3 - k^2 \) as \( (x - k)(x^2 + kx + k^2) \), and factor \( x^2 - k^2 \) as \( (x - k)(x + k) \). Thus, the equation becomes: \[ \frac{(x - k)(x^2 + kx + k^2)}{(x - k)(x + k)} \] Cancel the common factor \( x - k \): \[ \frac{x^2 + kx + k^2}{x + k} \] Now substitute \( x = k \): \[ \frac{k^2 + k^2 + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2} \] So, the right-hand side equals \( \frac{3k}{2} \). Step 3: Now, equate the two sides: \[ 4 = \frac{3k}{2} \] Solving for \( k \): \[ k = \frac{8}{3} \] Thus, the correct answer is (a) \( \frac{8}{3} \).