Calcule o limite:

j) Lim (x → 2) (x² + 5x - 14) / (x - 2)
k) Lim (x → 1) (x³ - x² + x - 1) / (x - 1)
l) Lim (x → -2) (x³ + 8) / (x + 2)
m) Lim (x → 5) (x² - 10x + _) / (x - 5)
n) Lim (x → ∞) (2x + 7)
o) Lim (x → ∞) (8x + 1) / (4x - 5)
p) Lim (x → ∞) (3x + 2) / (x² - 5x + 6)
q) Lim (x → ∞) (-2x² + 7) / (6x² + 1)

Q: Calcule o limite: j) Lim (x → 2) (x² + 5x - 14) / (x - 2) k) Lim (x → 1) (x³ - x² + x - 1) / (x - 1) l) Lim (x → -2) (x³ + 8) / (x + 2) m) Lim (x → 5) (x² - 10x + _) / (x - 5) n) Lim (x → ∞) (2x + 7) o) Lim (x → ∞) (8x + 1) / (4x - 5) p) Lim (x → ∞) (3x + 2) / (x² - 5x + 6) q) Lim (x → ∞) (-2x² + 7) / (6x² + 1)

Vamos calcular os limites solicitados: --- **j)** \[ \lim_{x \to 2} \frac{x^2 + 5x - 14}{x - 2} \] Fatorando o numerador $x^2 + 5x - 14$: \[ x^2 + 5x - 14 = (x - 2)(x + 7) \] Substituindo no limite: \[ \lim_{x \to 2} \frac{(x - 2)(x + 7)}{x - 2} \] Cancelando $x - 2$: \[ \lim_{x \to 2} (x + 7) = 2 + 7 = 9 \] **Resposta: 9** --- **k)** \[ \lim_{x \to 1} \frac{x^3 - x^2 + x - 1}{x - 1} \] Fatorando o numerador $x^3 - x^2 + x - 1$ por $x - 1$: \[ x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1) \] Substituindo no limite: \[ \lim_{x \to 1} \frac{(x - 1)(x^2 + 1)}{x - 1} \] Cancelando $x - 1$: \[ \lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2 \] **Resposta: 2** --- **l)** \[ \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \] Fatorando $x^3 + 8$ como uma soma de cubos: \[ x^3 + 8 = (x + 2)(x^2 - 2x + 4) \] Substituindo no limite: \[ \lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{x + 2} \] Cancelando $x + 2$: \[ \lim_{x \to -2} (x^2 - 2x + 4) = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12 \] **Resposta: 12** --- **m)** \[ \lim_{x \to 5} \frac{x^2 - 10x}{x - 5} \] Fatorando $x^2 - 10x$: \[ x^2 - 10x = x(x - 10) \] Substituindo no limite: \[ \lim_{x \to 5} \frac{x(x - 10)}{x - 5} \] Dividindo $x - 10$ por $x - 5$: \[ x - 10 = (x - 5) - 5 \implies \lim_{x \to 5} x - 5 = 0 \] ---

Question

Calcule o limite:

j) Lim (x → 2) (x² + 5x - 14) / (x - 2)
k) Lim (x → 1) (x³ - x² + x - 1) / (x - 1)
l) Lim (x → -2) (x³ + 8) / (x + 2)
m) Lim (x → 5) (x² - 10x + _) / (x - 5)
n) Lim (x → ∞) (2x + 7)
o) Lim (x → ∞) (8x + 1) / (4x - 5)
p) Lim (x → ∞) (3x + 2) / (x² - 5x + 6)
q) Lim (x → ∞) (-2x² + 7) / (6x² + 1)

Accepted Answer

Vamos calcular os limites solicitados:

---

**j)**  
$$
\lim_{x \to 2} \frac{x^2 + 5x - 14}{x - 2}
$$

Fatorando o numerador $x^2 + 5x - 14$:  
$$
x^2 + 5x - 14 = (x - 2)(x + 7)
$$

Substituindo no limite:  
$$
\lim_{x \to 2} \frac{(x - 2)(x + 7)}{x - 2}
$$

Cancelando $x - 2$:  
$$
\lim_{x \to 2} (x + 7) = 2 + 7 = 9
$$

**Resposta: 9**

---

**k)**  
$$
\lim_{x \to 1} \frac{x^3 - x^2 + x - 1}{x - 1}
$$

Fatorando o numerador $x^3 - x^2 + x - 1$ por $x - 1$:  
$$
x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1)
$$

Substituindo no limite:  
$$
\lim_{x \to 1} \frac{(x - 1)(x^2 + 1)}{x - 1}
$$

Cancelando $x - 1$:  
$$
\lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2
$$

**Resposta: 2**

---

**l)**  
$$
\lim_{x \to -2} \frac{x^3 + 8}{x + 2}
$$

Fatorando $x^3 + 8$ como uma soma de cubos:  
$$
x^3 + 8 = (x + 2)(x^2 - 2x + 4)
$$

Substituindo no limite:  
$$
\lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{x + 2}
$$

Cancelando $x + 2$:  
$$
\lim_{x \to -2} (x^2 - 2x + 4) = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12
$$

**Resposta: 12**

---

**m)**  
$$
\lim_{x \to 5} \frac{x^2 - 10x}{x - 5}
$$

Fatorando $x^2 - 10x$:  
$$
x^2 - 10x = x(x - 10)
$$

Substituindo no limite:  
$$
\lim_{x \to 5} \frac{x(x - 10)}{x - 5}
$$

Dividindo $x - 10$ por $x - 5$:  
$$
x - 10 = (x - 5) - 5 \implies \lim_{x \to 5} x - 5 = 0
$$

---