Let's go through each question based on the given information that $\lim_{{x \to 1}} f(x) = 3$.

Question 1

To solve for: $\lim_{{x \to 1}} f(x)(x - 1)$

Since $\lim_{{x \to 1}} f(x) = 3$ and $\lim_{{x \to 1}} (x - 1) = 0$, we can apply the limit multiplication rule: $\lim_{{x \to 1}} f(x)(x - 1) = \lim_{{x \to 1}} f(x) \cdot \lim_{{x \to 1}} (x - 1) = 3 \cdot 0 = 0$

Answer for Question 1: C. 0

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Question 2

To solve for: $\lim_{{x \to 1}} \frac{f(x)}{2x - 5}$

Substitute $x = 1$ into $2x - 5$: $2 \cdot 1 - 5 = 2 - 5 = -3$

Thus, $\lim_{{x \to 1}} \frac{f(x)}{2x - 5} = \frac{\lim_{{x \to 1}} f(x)}{\lim_{{x \to 1}} (2x - 5)} = \frac{3}{-3} = -1$

Answer for Question 2: B. -1

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Question 3

To solve for: $\lim_{{x \to 1}} \frac{f(x) + x^2}{x + 1}$

First, calculate each limit separately:

1. $\lim_{{x \to 1}} f(x) = 3$
2. $\lim_{{x \to 1}} x^2 = 1^2 = 1$
3. $\lim_{{x \to 1}} (x + 1) = 1 + 1 = 2$

Thus, $\lim_{{x \to 1}} \frac{f(x) + x^2}{x + 1} = \frac{\lim_{{x \to 1}} f(x) + \lim_{{x \to 1}} x^2}{\lim_{{x \to 1}} (x + 1)} = \frac{3 + 1}{2} = \frac{4}{2} = 2$

Answer for Question 3: B. 2

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Summary of Answers

1. C. 0
2. B. -1
3. B. 2

Would you like more details or explanations for these solutions?

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Here are five related questions to deepen your understanding:

1. What are the general rules for calculating limits when a term approaches zero?
2. How can we apply L'Hôpital's Rule in indeterminate forms involving division?
3. How does continuity affect the limits of composed functions?
4. Why is it valid to separate limits in multiplication or division, given that the individual limits exist?
5. How would the answers change if $\lim_{{x \to 1}} f(x)$ were a different value?

Tip: Always check if a function is continuous at the point of interest, as this can often simplify limit calculations.