Let's evaluate the limit [tex]\(\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1}\)[/tex]. To do so, we'll follow a step-by-step approach:
1. Rewrite the numerator and the denominator in factored form:
The numerator [tex]\(x^2 - 1\)[/tex] can be factored using the difference of squares:
[tex]\[
x^2 - 1 = (x - 1)(x + 1)
\][/tex]
The denominator [tex]\(x^3 - 1\)[/tex] can be factored using the difference of cubes:
[tex]\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\][/tex]
2. Substitute the factored forms into the limit expression:
[tex]\[
\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)}
\][/tex]
3. Simplify the fraction by canceling common factors:
Notice that the factor [tex]\((x - 1)\)[/tex] appears in both the numerator and the denominator. Assuming [tex]\(x \neq 1\)[/tex], we can cancel [tex]\((x - 1)\)[/tex]:
[tex]\[
\lim_{x \to 1} \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)} = \lim_{x \to 1} \frac{x + 1}{x^2 + x + 1}
\][/tex]
4. Evaluate the simplified limit by direct substitution:
Substitute [tex]\(x = 1\)[/tex] into the simplified expression:
[tex]\[
\frac{x + 1}{x^2 + x + 1}\bigg|_{x = 1} = \frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{3}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1} = \frac{2}{3}
\][/tex]
