Question

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

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

Hint:

If direct substitution gives \(\frac{0}{0}\), factorize the expression and cancel the common factor before applying the limit.

Answer 1

The Correct Option is C

We need to evaluate: \[ \lim_{x\to 1}\frac{x^3-1}{x-1}. \] If we directly substitute \(x=1\), then: \[ \frac{1^3-1}{1-1}=\frac{0}{0}. \] This is an indeterminate form. So, we factorize the numerator. Using the identity: \[ a^3-b^3=(a-b)(a^2+ab+b^2), \] we get \[ x^3-1=(x-1)(x^2+x+1). \] Therefore, \[ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1}. \] Cancel the common factor \((x-1)\): \[ = x^2+x+1. \] Now substitute \(x=1\): \[ 1^2+1+1=3. \] Hence, \[ \lim_{x\to 1}\frac{x^3-1}{x-1}=3. \]