Step 1: Initial Assessment:
The limit presents an indeterminate form of type \(\frac{0}{0}\) since direct substitution of \(x=1\) results in \(\frac{1^3-1}{1-1} = \frac{0}{0}\). This can be resolved by factoring or by applying L'Hôpital's Rule.
Step 2: Methods:
Method 1: Factoring
Apply the difference of cubes factorization: \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\).
Method 2: L'Hôpital's Rule
For limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided the limit on the right exists.
Step 3: Solution:
Factoring Approach:
Factor \(x^3-1\) as \(x^3 - 1^3\):\[ x^3 - 1 = (x-1)(x^2 + x . 1 + 1^2) = (x-1)(x^2+x+1) \]Substitute the factored form back into the original limit:\[ \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{x-1} \]Cancel the common factor \((x-1)\), valid for \(x eq 1\):\[ \lim_{x \to 1} (x^2+x+1) \]Evaluate the limit by substituting \(x=1\):\[ 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \]L'Hôpital's Rule Approach:
Define \(f(x) = x^3 - 1\) and \(g(x) = x - 1\).
Compute the derivatives: \(f'(x) = 3x^2\) and \(g'(x) = 1\).
Apply L'Hôpital's Rule:\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{3x^2}{1} \]Evaluate by substituting \(x=1\):\[ \frac{3(1)^2}{1} = 3 \]
Step 4: Conclusion:
Both methods converge to the same solution. Therefore, the limit is 3.