Answer :

To determine the limit [tex]\(\lim_{x \rightarrow 1} \frac{x^2 - 5x + 4}{x^2 + 6x - 7}\)[/tex], let's follow a step-by-step approach.

1. Factoring the Numerator and Denominator:
First, we simplify the expression by factoring both the numerator and the denominator.

- For the numerator [tex]\(x^2 - 5x + 4\)[/tex]:

This quadratic expression can be factored into:
[tex]\[ x^2 - 5x + 4 = (x - 1)(x - 4) \][/tex]

- For the denominator [tex]\(x^2 + 6x - 7\)[/tex]:

This quadratic expression can be factored into:
[tex]\[ x^2 + 6x - 7 = (x + 7)(x - 1) \][/tex]

2. Substituting the Factored Forms:
Substituting the factored forms into the original expression, we get:
[tex]\[ \frac{x^2 - 5x + 4}{x^2 + 6x - 7} = \frac{(x - 1)(x - 4)}{(x + 7)(x - 1)} \][/tex]

3. Canceling Common Factors:
Notice that [tex]\((x - 1)\)[/tex] is a common factor in both the numerator and the denominator. As long as [tex]\(x \neq 1\)[/tex], we can cancel these factors:
[tex]\[ \frac{(x - 1)(x - 4)}{(x + 7)(x - 1)} = \frac{x - 4}{x + 7} \][/tex]

4. Taking the Limit:
Now, we take the limit of the simplified expression as [tex]\(x\)[/tex] approaches 1:
[tex]\[ \lim_{x \rightarrow 1} \frac{x - 4}{x + 7} \][/tex]
Substitute [tex]\( x = 1 \)[/tex] into [tex]\(\frac{x - 4}{x + 7}\)[/tex]:
[tex]\[ \frac{1 - 4}{1 + 7} = \frac{-3}{8} \][/tex]

Hence, the limit is:
[tex]\[ \boxed{\frac{-3}{8}} \][/tex]