Question

If \( \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} \), then the value of \( k \) is:

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

Hint:

For limits of type \( \frac{f(x)-f(a)}{x-a} \), factorization helps simplify directlyAlways cancel common factors before substitution.

Answer 1

The Correct Option is D

Step 1: Evaluate first limit.
\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \]
Factor numerator:
\[ x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1) \]
So:
\[ = \lim_{x \to 1} (x^3 + x^2 + x + 1) \]

Step 2: Substitute \( x = 1 \).
\[ = 1 + 1 + 1 + 1 = 4 \]

Step 3: Evaluate second limit.
\[ \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} \]
Factor both numerator and denominator:
\[ x^3 - k^3 = (x - k)(x^2 + xk + k^2) \]
\[ x^2 - k^2 = (x - k)(x + k) \]

Step 4: Simplify expression.
\[ = \lim_{x \to k} \frac{x^2 + xk + k^2}{x + k} \]

Step 5: Substitute \( x = k \).
\[ = \frac{k^2 + k^2 + k^2}{2k} \]
\[ = \frac{3k^2}{2k} = \frac{3k}{2} \]

Step 6: Equate both limits.
\[ 4 = \frac{3k}{2} \]

Step 7: Solve for \( k \).
\[ k = \frac{8}{3} \]
\[ \boxed{\frac{8}{3}} \]