1.2.5. a) $$ \lim _{n \rightarrow \infty} \frac{(6 n)^{2}-(6+n)^{2}}{(6+n)^{2}-(1-n)^{2}} ; $$ b) $$ \lim _{n \rightarrow \infty} \frac{\sqrt{n^{2}-8}-n \sqrt{n\left(n^{2}+5\right)}}{\sqrt{n}} $$; v) $$ \lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2}{2 n^{2}+1}\right)^{n^{2}-1} $$

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Let's evaluate each of the given limits step by step.

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### **Problem 1.2.5.a**

Evaluate:
$$
\lim_{n \to \infty} \frac{(6n)^2 - (6 + n)^2}{(6 + n)^2 - (1 - n)^2}
$$

**Step 1: Expand the Numerator and Denominator**

- **Numerator:**
  $$
  (6n)^2 - (6 + n)^2 = 36n^2 - (n^2 + 12n + 36) = 35n^2 - 12n - 36
  $$

- **Denominator:**
  $$
  (6 + n)^2 - (1 - n)^2 = (n^2 + 12n + 36) - (n^2 - 2n + 1) = 14n + 35
  $$

**Step 2: Simplify the Limit**

$$
\lim_{n \to \infty} \frac{35n^2 - 12n - 36}{14n + 35}
$$

As $$n \to \infty$$, the highest degree terms dominate:

$$
\frac{35n^2}{14n} = \frac{35}{14}n = \frac{5}{2}n \to \infty
$$

**Answer:** The limit is $$ +\infty $$.

---

### **Problem 1.2.5.b**

Evaluate:
$$
\lim_{n \to \infty} \frac{\sqrt{n^2 - 8} - n \sqrt{n(n^2 + 5)}}{\sqrt{n}}
$$

**Step 1: Approximate Each Term for Large $$n$$**

- **First Term:**
  $$
  \sqrt{n^2 - 8} \approx n - \frac{4}{n} \quad \text{(using the expansion } \sqrt{n^2 - a} \approx n - \frac{a}{2n})
  $$

- **Second Term:**
  $$
  n \sqrt{n(n^2 + 5)} = n \sqrt{n^3 + 5n} = n \cdot n^{3/2} \sqrt{1 + \frac{5}{n^2}} \approx n^{5/2} \left(1 + \frac{5}{2n^2}\right)
  $$

**Step 2: Substitute the Approximations into the Limit**

$$
\frac{\sqrt{n^2 - 8} - n \sqrt{n(n^2 + 5)}}{\sqrt{n}} \approx \frac{n - \frac{4}{n} - \left(n^{5/2} + \frac{5}{2}n^{1/2}\right)}{\sqrt{n}} = \frac{-n^{5/2} + n - \frac{5}{2}n^{1/2} - \frac{4}{n}}{\sqrt{n}}
$$

Simplify each term:

$$
\frac{-n^{5/2}}{\sqrt{n}} = -n^2, \quad \frac{n}{\sqrt{n}} = n^{1/2}, \quad \frac{-\frac{5}{2}n^{1/2}}{\sqrt{n}} = -\frac{5}{2}, \quad \frac{-\frac{4}{n}}{\sqrt{n}} = -\frac{4}{n^{3/2}}
$$

So,

$$
\approx -n^2 + n^{1/2} - \frac{5}{2} - \frac{4}{n^{3/2}}
$$

**Step 3: Determine the Dominant Term**

As $$n \to \infty$$, the $$ -n^2 $$ term dominates.

**Answer:** The limit is $$ -\infty $$.

---

### **Problem 1.2.5.v**

Evaluate:
$$
\lim_{n \to \infty} \left(\frac{2n^2 + 2}{2n^2 + 1}\right)^{n^2 - 1}
$$

**Step 1: Simplify the Base**

$$
\frac{2n^2 + 2}{2n^2 + 1} = \frac{2n^2(1 + \frac{1}{n^2})}{2n^2(1 + \frac{1}{2n^2})} = \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{2n^2}} \approx 1 + \frac{1}{2n^2} \quad \text{for large } n
$$

**Step 2: Express the Limit Using the Natural Exponential Function**

Recognize that:
$$
\left(1 + \frac{a}{n^2}\right)^{n^2} \approx e^{a} \quad \text{as } n \to \infty
$$

Here, $$ a = \frac{1}{2} $$, and the exponent is $$n^2 - 1 \approx n^2$$.

**Step 3: Apply the Approximation**

$$
\left(1 + \frac{1}{2n^2}\right)^{n^2 - 1} \approx e^{\frac{1}{2}} = \sqrt{e}
$$

**Answer:** The limit is $$ \sqrt{e} $$.

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### **Summary of Answers**

1. **a)** $$ +\infty $$

2. **b)** $$ -\infty $$

3. **v)** $$ \sqrt{e} $$
The answers are: a) $$ +\infty $$ b) $$ -\infty $$ v) $$ \sqrt{e} $$

1.2.5. a) b) ; v)

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Solución

Problem 1.2.5.a

Problem 1.2.5.b

Problem 1.2.5.v

Summary of Answers

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