Let's evaluate these limits using the limit laws. I'll go through each part one-by-one:

(i)

$\lim_{{x \to 1}} \frac{3x^2 + \sqrt{x} + 5}{x^3 + 2}$ To evaluate this, we can substitute $x = 1$ directly, as this is a continuous function and there's no indication of division by zero.

1. Substitute $x = 1$: $\frac{3(1)^2 + \sqrt{1} + 5}{(1)^3 + 2} = \frac{3 + 1 + 5}{1 + 2} = \frac{9}{3} = 3$

So, the result for (i) is: $\lim_{{x \to 1}} \frac{3x^2 + \sqrt{x} + 5}{x^3 + 2} = 3$

(ii)

$\lim_{{x \to 1}} \frac{x^2 + 2x - 3}{x^2 - 1}$ In this case, direct substitution would lead to a $\frac{0}{0}$ indeterminate form, so we should try factoring.

1. Factor the numerator and the denominator: $x^2 + 2x - 3 = (x + 3)(x - 1)$ $x^2 - 1 = (x + 1)(x - 1)$

2. Rewrite the expression by canceling out the $(x - 1)$ term: $\frac{(x + 3)(x - 1)}{(x + 1)(x - 1)} = \frac{x + 3}{x + 1}$

3. Now substitute $x = 1$: $\frac{1 + 3}{1 + 1} = \frac{4}{2} = 2$

So, the result for (ii) is: $\lim_{{x \to 1}} \frac{x^2 + 2x - 3}{x^2 - 1} = 2$

(iii)

$\lim_{{x \to 2^+}} \frac{x^2 - 4}{|x - 2|}$ Here, we need to analyze the behavior of $x$ as it approaches 2 from the right ($x \to 2^+$).

1. Simplify $x^2 - 4$ as a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$ So the expression becomes: $\frac{(x - 2)(x + 2)}{|x - 2|}$

2. For $x \to 2^+$, we have $x - 2 > 0$, so $|x - 2| = x - 2$. This allows us to cancel out $(x - 2)$: $\frac{(x - 2)(x + 2)}{x - 2} = x + 2$

3. Now, substitute $x = 2$: $x + 2 = 2 + 2 = 4$

Thus, the result for (iii) is: $\lim_{{x \to 2^+}} \frac{x^2 - 4}{|x - 2|} = 4$

Summary of Results

1. $(i)$ $\lim_{{x \to 1}} \frac{3x^2 + \sqrt{x} + 5}{x^3 + 2} = 3$
2. $(ii)$ $\lim_{{x \to 1}} \frac{x^2 + 2x - 3}{x^2 - 1} = 2$
3. $(iii)$ $\lim_{{x \to 2^+}} \frac{x^2 - 4}{|x - 2|} = 4$

Would you like further explanations on any of these parts?

Here are some related questions to expand your understanding:

1. How would the results change if we approached from the left in part (iii)?
2. What happens if we use L'Hôpital's Rule for part (ii)?
3. Can limit laws be applied to more complex functions?
4. Why is direct substitution sometimes insufficient in limit calculations?
5. What other techniques can resolve indeterminate forms beyond factoring?

Tip: Always check if a function can be simplified before applying L'Hôpital’s Rule—it can save time and avoid unnecessary differentiation.