The problem you provided asks to find the limit:

$\lim_{{x \to 1}} \frac{{x^2 - 5x + 4}}{{x^2 - 1}}$

Step-by-step solution:

1. Substitute $x = 1$ directly into the function to check if it results in an indeterminate form:

$\frac{{1^2 - 5(1) + 4}}{{1^2 - 1}} = \frac{{1 - 5 + 4}}{{1 - 1}} = \frac{0}{0}$

This results in the indeterminate form $\frac{0}{0}$, which means we can apply factoring to simplify the expression before calculating the limit.

2. Factor the numerator: $x^2 - 5x + 4$

   • This is a quadratic expression, and it can be factored as:

   $x^2 - 5x + 4 = (x - 4)(x - 1)$

3. Factor the denominator: $x^2 - 1$

   • This is a difference of squares, which can be factored as:

   $x^2 - 1 = (x - 1)(x + 1)$

4. Simplify the expression: Now substitute the factored forms back into the limit:

   $\frac{{(x - 4)(x - 1)}}{{(x - 1)(x + 1)}}$

   The $(x - 1)$ terms cancel out (since $x \neq 1$ in the limit process):

   $\lim_{{x \to 1}} \frac{{x - 4}}{{x + 1}}$

5. Substitute $x = 1$ into the simplified expression:

   $\frac{{1 - 4}}{{1 + 1}} = \frac{-3}{2}$

Thus, the limit is:

$\boxed{-\frac{3}{2}}$

Would you like further details or explanations? Here are some related questions to deepen your understanding:

1. What are other methods to solve limits besides factoring?
2. What is L'Hôpital's Rule, and how would it apply to indeterminate forms like $\frac{0}{0}$?
3. How do you solve limits involving complex functions?
4. Can limits exist if a function is discontinuous at a point?
5. How does the difference of squares factoring technique work?

Tip: Always check for simple factoring opportunities when faced with indeterminate forms like $\frac{0}{0}$ in limits.