To solve the limit [tex]\(\lim_{x \to 1} \frac{x^2 + 2x - 1}{x + 1}\)[/tex], follow these steps:
1. Substitute [tex]\( x = 1 \)[/tex] into the function: We start by directly substituting [tex]\( x \)[/tex] with 1 to see if it simplifies easily.
[tex]\[
f(1) = \frac{1^2 + 2 \cdot 1 - 1}{1 + 1} = \frac{1 + 2 - 1}{2} = \frac{2}{2} = 1
\][/tex]
Here, substituting [tex]\( x = 1 \)[/tex] gives a result without any indeterminate form.
2. Simplify the expression algebraically: In general, simplifying the expression before substituting can be a good practice:
[tex]\[
\frac{x^2 + 2x - 1}{x + 1}
\][/tex]
However, in this particular case, the direct substitution already provided a clear and simple value. There is no indeterminate form like [tex]\( \frac{0}{0} \)[/tex] when [tex]\( x = 1 \)[/tex], so further factorization or simplification isn't necessary.
3. Consider factoring (additional step for more complex limits): If there were an indeterminate form, we would consider factoring. We can suppose the polynomial in the numerator can be factored for further simplification:
[tex]\[
x^2 + 2x - 1 = (x + 1)(x - 1)
\][/tex]
Then the expression becomes:
[tex]\[
\frac{(x + 1)(x - 1)}{x + 1} = x - 1 \quad \text{for } x \neq -1
\][/tex]
However, since [tex]\( x = 1 \)[/tex], the immediate substitution already validated our solution.
4. Conclusion: Since direct substitution and simplification yielded a straightforward value:
Therefore, the limit is:
[tex]\[
\lim_{x \to 1} \frac{x^2 + 2x - 1}{x + 1} = 1.
\][/tex]
