2) \( \lim _{x \rightarrow+\infty} \frac{x^{3}+3}{\sqrt{x+4}} \) 3) \( \lim _{x \rightarrow-\infty} \frac{1-2 x}{x^{3}+2} \) 4) \( \lim _{x \rightarrow-3} \frac{\sqrt{x+3}}{x^{2}-9} \)

We will evaluate each limit step‐by‐step. ––––––––––––––– Problem 2: limₓ→+∞ (x³ + 3) / √(x + 4) For very large x, the highest‐power term dominates in the numerator and denominator: • The numerator is approximately x³. • In the denominator, since √(x + 4) ~ √x, the behavior is like √x. Thus, the expression behaves like   x³ / √x = x^(3 – 1/2) = x^(5/2). Since x^(5/2) → +∞ as x → +∞, the limit diverges to +∞. ––––––––––––––– Problem 3: limₓ→–∞ (1 – 2x) / (x³ + 2) For large negative x, the dominant powers are: • In the numerator, the term –2x dominates. • In the denominator, the term x³ dominates. Thus the limit is approximately   (–2x) / x³ = –2 / x². Since x² becomes very large (always positive) as x → –∞, –2/x² → 0. (Notice that although –2/x² is negative, its magnitude tends to 0.) Therefore, the limit is 0. ––––––––––––––– Problem 4: limₓ→–3 (√(x + 3)) / (x² – 9) Note that √(x + 3) is defined for x + 3 ≥ 0, so we consider x approaching –3 from the right. First, factor the denominator:   x² – 9 = (x + 3)(x – 3). Then we have   (√(x + 3))/( (x + 3)(x – 3) ). Write the fraction as   [1/(x – 3)] · [√(x + 3)/(x + 3)]. Notice that √(x + 3)/(x + 3) = 1/√(x + 3), so the expression becomes   1/[(x – 3) √(x + 3)]. Now, examine the behavior as x → –3⁺: • √(x + 3) → 0⁺. • x – 3 → –3 – 3 = –6. Thus, near x = –3 the expression behaves like   1/[(–6)·(√(x + 3))] = –1/(6√(x + 3)). Since √(x + 3) tends to 0 from the positive side, 1/√(x + 3) blows up to +∞. Multiplying by –1/6 gives –∞. ––––––––––––––– Final Answers: 2) The limit is +∞. 3) The limit is 0. 4) The limit is –∞.