The problem involves evaluating the one-sided limit:

$\lim_{x \to 4^+} \frac{x^3 + 2x^2 - 3x}{x^2 - 5x + 4}.$

Step-by-Step Solution:

1. Factorize the denominator: $x^2 - 5x + 4 = (x - 4)(x - 1).$

2. Analyze the numerator: The numerator is $x^3 + 2x^2 - 3x$. Factor out $x$: $x^3 + 2x^2 - 3x = x(x^2 + 2x - 3).$ Now factorize $x^2 + 2x - 3$: $x^2 + 2x - 3 = (x + 3)(x - 1).$ Therefore, the numerator becomes: $x(x + 3)(x - 1).$

3. Simplify the fraction: Substituting the factored forms of the numerator and denominator: $\frac{x^3 + 2x^2 - 3x}{x^2 - 5x + 4} = \frac{x(x + 3)(x - 1)}{(x - 4)(x - 1)}.$ Cancel out the $(x - 1)$ term (valid since $x \neq 1$): $\frac{x(x + 3)}{x - 4}.$

4. Evaluate the one-sided limit as $x \to 4^+$: Substitute $x = 4$ into the simplified fraction: $\frac{x(x + 3)}{x - 4} = \frac{4(4 + 3)}{4 - 4}.$ This results in: $\frac{4 \cdot 7}{0^+} = \frac{28}{0^+}.$

   Since the denominator approaches $0^+$ (positive from the right-hand side), the fraction tends towards $+\infty$.

Final Answer:

$\lim_{x \to 4^+} \frac{x^3 + 2x^2 - 3x}{x^2 - 5x + 4} = +\infty.$

Would you like further explanation or more examples of similar problems? Here are 5 related questions to consider:

1. What is the significance of one-sided limits, and how do they differ from two-sided limits?
2. How do you handle limits involving indeterminate forms, such as $0/0$?
3. Can this method of factoring and simplifying always be applied for polynomial expressions?
4. How does the behavior of a function change near a vertical asymptote?
5. What is the role of limits in determining the continuity of a function?

Tip:

When evaluating limits, always check for factors that cancel in the numerator and denominator, as they may reveal removable discontinuities or simplify the problem.