To find the limit [tex]\(\lim_{x \rightarrow 1} \frac{x^2-1}{x^3-1}\)[/tex], we need to carefully examine the behavior of the function as the variable [tex]\(x\)[/tex] approaches 1. Let's evaluate this step by step.
1. Factorize the numerator and the denominator:
The given function is [tex]\(\frac{x^2-1}{x^3-1}\)[/tex].
First, factorize [tex]\(x^2-1\)[/tex]:
[tex]\[
x^2 - 1 = (x - 1)(x + 1)
\][/tex]
Next, factorize [tex]\(x^3-1\)[/tex]:
[tex]\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\][/tex]
2. Rewrite the function with these factorizations:
[tex]\[
\frac{x^2 - 1}{x^3 - 1} = \frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)}
\][/tex]
3. Simplify the function by canceling out common factors:
Notice that [tex]\((x-1)\)[/tex] is a common factor in both the numerator and the denominator. Thus, we can cancel out [tex]\((x-1)\)[/tex]:
[tex]\[
\frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x + 1)} = \frac{x + 1}{x^2 + x + 1}
\][/tex]
provided [tex]\(x \neq 1\)[/tex].
4. Substitute [tex]\(x = 1\)[/tex] into the simplified function:
Now, substitute [tex]\(x = 1\)[/tex] into the simplified function to find:
[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]\[
\lim_{x \rightarrow 1} \frac{x^2 - 1}{x^3 - 1} = \frac{2}{3}
\][/tex]
