To determine the limit
[tex]\[ \lim_{x \to -1} \frac{x - 2}{x^2 + 4x - 3}, \][/tex]
we first analyze the function [tex]\(\frac{x - 2}{x^2 + 4x - 3}\)[/tex].
1. Factor the Denominator:
The quadratic expression in the denominator can be factored to make it easier to work with:
[tex]\[
x^2 + 4x - 3 = (x + 5)(x - 1).
\][/tex]
However, upon simplifying further, we can rewrite the correct factors as:
[tex]\[
x^2 + 4x - 3 = (x + 5)(x - 1).
\][/tex]
Note, though, that factoring this correctly gives:
[tex]\[
x^2 + 4x - 3 = (x + 5)(x - 1).
\][/tex]
2. Substitution Approach:
Normally, after obtaining factored forms, we could check values directly by substituting [tex]\( x = -1 \)[/tex] into the expression. But in this case, it helps to directly substitute instead:
[tex]\[
\frac{(-1) - 2}{(-1)^2 + 4(-1) - 3}.
\][/tex]
3. Computation:
Substituting [tex]\( x = -1 \)[/tex] into the equation yields:
[tex]\[
\frac{-1 - 2}{1 + 4(-1) - 3}.
\][/tex]
Simplifying the values step-by-step:
[tex]\[
\frac{-3}{1 - 4 - 3} = \frac{-3}{-6}.
\][/tex]
So,
[tex]\[
\frac{-3}{-6} = \frac{1}{2}.
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \to -1} \frac{x - 2}{x^2 + 4x - 3} = \frac{1}{2}.
\][/tex]
