Question

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

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

Hint:

Use \(x^3-1=(x-1)(x^2+x+1)\) for limits involving \(x\to 1\).

Answer 1

The Correct Option is C

Step 1: Factorize: \[ x^3-1=(x-1)(x^2+x+1) \]

Step 2: \[ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1} \]

Step 3: Cancel \((x-1)\): \[ =x^2+x+1 \]

Step 4: Put \(x=1\): \[ 1^2+1+1=3 \] \[ \boxed{3} \]