The limit expression given in the image is:

$\lim_{x \to 1^+} \left( \frac{9}{x^2 - x} - \frac{x}{x - 1} \right)$

Let's break this down and solve step by step.

Step 1: Simplify each term

• The first term is $\frac{9}{x^2 - x}$, which can be factored: $x^2 - x = x(x - 1)$ So the first term becomes: $\frac{9}{x(x - 1)}$

• The second term is $\frac{x}{x - 1}$.

Step 2: Combine the terms

Both terms have the denominator $x - 1$, so let's combine them: $\frac{9}{x(x - 1)} - \frac{x}{x - 1} = \frac{9 - x^2}{x(x - 1)}$ The numerator $9 - x^2$ can be rewritten as: $9 - x^2 = (3 - x)(3 + x)$ Thus, the expression becomes: $\frac{(3 - x)(3 + x)}{x(x - 1)}$

Step 3: Evaluate the limit as $x \to 1^+$

As $x$ approaches 1 from the right, we need to check if there are any issues with division by zero. Notice that the denominator contains $x(x - 1)$, and as $x \to 1^+$, $x - 1$ approaches 0. This suggests that the limit might involve an infinite behavior.

Let’s analyze each part:

• As $x \to 1^+$, the factor $x - 1$ in the denominator becomes very small and positive, so the whole expression tends towards infinity.

Thus, the limit tends to $-\infty$ as $x \to 1^+$, because the numerator $(3 - x)(3 + x)$ evaluates to 4 when $x = 1$, but the denominator tends to 0 from the positive side.

Final Answer:

$\lim_{x \to 1^+} \left( \frac{9}{x^2 - x} - \frac{x}{x - 1} \right) = -\infty$

Would you like further details or have any other questions?

Here are five related questions to explore this topic further:

1. What happens if you evaluate the limit as $x \to 1^-$?
2. How can L'Hôpital's Rule be applied to similar limit problems?
3. What does it mean for a limit to approach infinity or negative infinity?
4. How would the limit change if the numerator in one of the fractions was modified?
5. How can factoring be useful in solving complex limit problems?

Tip: Always check for indeterminate forms when calculating limits involving fractions, especially if the denominator tends to zero.