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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.

This problem involves finding the limit of the rational function (x^2 - 5x + 4) / (x^2 - 1) as x approaches 1. It includes the use of factoring quadratic equations and simplifying indeterminate forms. A clear step-by-step solution for students in Grades 11-12.

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