To find the limit [tex]\(\lim _{n \rightarrow 1} \frac{n^3-1}{n^2-1}\)[/tex], we need to simplify the expression and then evaluate it as [tex]\(n\)[/tex] approaches 1.
### Step-by-Step Solution
1. Factor the numerator and the denominator:
- The numerator [tex]\(n^3 - 1\)[/tex] can be factored using the difference of cubes formula:
[tex]\[
n^3 - 1 = (n - 1)(n^2 + n + 1)
\][/tex]
- The denominator [tex]\(n^2 - 1\)[/tex] can be factored using the difference of squares formula:
[tex]\[
n^2 - 1 = (n - 1)(n + 1)
\][/tex]
2. Substitute the factored forms:
[tex]\[
\frac{n^3 - 1}{n^2 - 1} = \frac{(n - 1)(n^2 + n + 1)}{(n - 1)(n + 1)}
\][/tex]
3. Cancel the common factor [tex]\((n - 1)\)[/tex]:
Since [tex]\(n \neq 1\)[/tex], the [tex]\((n - 1)\)[/tex] terms in the numerator and denominator can be canceled out:
[tex]\[
\frac{(n - 1)(n^2 + n + 1)}{(n - 1)(n + 1)} = \frac{n^2 + n + 1}{n + 1}
\][/tex]
4. Simplify the resulting expression:
Now, we have the simplified expression:
[tex]\[
\frac{n^2 + n + 1}{n + 1}
\][/tex]
5. Evaluate the simplified expression as [tex]\(n\)[/tex] approaches 1:
Substitute [tex]\(n = 1\)[/tex] into the simplified expression:
- Numerator:
[tex]\[
1^2 + 1 + 1 = 3
\][/tex]
- Denominator:
[tex]\[
1 + 1 = 2
\][/tex]
Therefore, the expression evaluated at [tex]\(n = 1\)[/tex] is:
[tex]\[
\frac{3}{2} = 1.5
\][/tex]
### Conclusion
The limit is:
[tex]\[
\lim _{n \rightarrow 1} \frac{n^3-1}{n^2-1} = 1.5
\][/tex]
The detailed step-by-step solution shows that the limit of the given expression as [tex]\(n\)[/tex] approaches 1 is [tex]\(\boxed{1.5}\)[/tex].
