Sure, let's solve the limit \(\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}\) step-by-step.
1. Identify the given expression:
We need to find the limit of \(\frac{x^2 - 1}{x - 1}\) as \(x\) approaches 1.
2. Factorize the numerator:
Notice that the numerator \(x^2 - 1\) is a difference of squares. This can be factorized as:
[tex]\[
x^2 - 1 = (x + 1)(x - 1)
\][/tex]
Now our expression becomes:
[tex]\[
\frac{(x + 1)(x - 1)}{x - 1}
\][/tex]
3. Simplify the expression:
Notice that \((x - 1)\) appears in both the numerator and the denominator. As long as \(x \neq 1\), we can cancel out the \((x - 1)\) term:
[tex]\[
\frac{(x + 1)(x - 1)}{x - 1} = x + 1
\][/tex]
4. Evaluate the limit:
Now we have a much simpler expression to evaluate:
[tex]\[
\lim_{x \rightarrow 1} (x + 1)
\][/tex]
As \(x\) approaches 1, we substitute \(x = 1\) into the simplified expression:
[tex]\[
x + 1 = 1 + 1 = 2
\][/tex]
5. Conclusion:
Therefore, the limit is:
[tex]\[
\lim_{x \rightarrow 1} \frac{x^2-1}{x-1} = 2
\][/tex]
Hence, the solution is [tex]\(\boxed{2}\)[/tex].
