To evaluate the limit $\lim_{x \rightarrow 1} \frac{x^2 + 2x - 3}{x^2 - 5x + 4}$, we first check the direct substitution:

$\frac{1^2 + 2 \cdot 1 - 3}{1^2 - 5 \cdot 1 + 4} = \frac{1 + 2 - 3}{1 - 5 + 4} = \frac{0}{0}$

Since direct substitution gives us the indeterminate form $\frac{0}{0}$, we need to simplify the expression by factoring the numerator and the denominator.

The numerator $x^2 + 2x - 3$ factors as $(x + 3)(x - 1)$.

The denominator $x^2 - 5x + 4$ factors as $(x - 4)(x - 1)$.

So, we have:

$\frac{x^2 + 2x - 3}{x^2 - 5x + 4} = \frac{(x + 3)(x - 1)}{(x - 4)(x - 1)}$

We can cancel the common factor $(x - 1)$ in the numerator and the denominator, provided $x \neq 1$:

$\frac{(x + 3)(x - 1)}{(x - 4)(x - 1)} = \frac{x + 3}{x - 4} \quad \text{for} \quad x \neq 1$

Now, we can evaluate the limit as $x$ approaches 1:

$\lim_{x \rightarrow 1} \frac{x + 3}{x - 4} = \frac{1 + 3}{1 - 4} = \frac{4}{-3} = -\frac{4}{3}$

Therefore,

$\lim_{x \rightarrow 1} \frac{x^2 + 2x - 3}{x^2 - 5x + 4} = -\frac{4}{3}$

Would you like more details or have any questions about this solution? Here are five more questions you might be interested in:

1. How do you find limits involving indeterminate forms?
2. What are the common techniques for factoring polynomials?
3. How do you apply L'Hôpital's rule to solve limits?
4. What are the different types of indeterminate forms?
5. How do you solve limits involving trigonometric functions?

Tip: When you encounter an indeterminate form like $\frac{0}{0}$ in limits, try simplifying the expression by factoring, canceling common terms, or applying L'Hôpital's rule if appropriate.