you apply the Rule of L'Hopital. Enter the letter "D" if the $$ \lim _{x \rightarrow 0} \frac{\sin x}{x}=\square $$. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}}=\square $$. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin (x)}=\square $$. $$ \lim _{x \rightarrow-2} \frac{x^{2}+5 x+6}{x-2}=\square $$. $$ \lim _{x \rightarrow 2} \frac{x^{2}+5 x+6}{x-2}=\square $$.

Question

UpStudy AI · Accepted Answer

Sure, let's evaluate each of the given limits using **L'Hôpital's Rule** where applicable.

### 1. $$ \lim _{x \rightarrow 0} \frac{\sin x}{x} $$

**Evaluation:**

As $$ x \rightarrow 0 $$,
- Numerator $$ \sin x \rightarrow 0 $$
- Denominator $$ x \rightarrow 0 $$

This results in the indeterminate form $$ \frac{0}{0} $$. Applying L'Hôpital's Rule:

$$
\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1
$$

**Result:**  
$$ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $$

---

### 2. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} $$

**Evaluation:**

As $$ x \rightarrow 0 $$,
- Numerator $$ e^{x} - 1 - x \rightarrow 0 $$
- Denominator $$ x^2 \rightarrow 0 $$

Indeterminate form $$ \frac{0}{0} $$. Applying L'Hôpital's Rule twice:

**First Application:**
$$
\lim_{x \to 0} \frac{e^{x} - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^{x} - 1}{2x}
$$

**Second Application:**
$$
\lim_{x \to 0} \frac{e^{x} - 1}{2x} = \lim_{x \to 0} \frac{e^{x}}{2} = \frac{1}{2}
$$

**Result:**  
$$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} = \frac{1}{2} $$

---

### 3. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin (x)} $$

**Evaluation:**

As $$ x \rightarrow 0 $$,
- Numerator $$ e^{x} - e^{-x} \rightarrow 0 $$
- Denominator $$ \sin(x) \rightarrow 0 $$

Indeterminate form $$ \frac{0}{0} $$. Applying L'Hôpital's Rule:

$$
\lim_{x \to 0} \frac{e^{x} - e^{-x}}{\sin(x)} = \lim_{x \to 0} \frac{e^{x} + e^{-x}}{\cos(x)} = \frac{1 + 1}{1} = 2
$$

**Result:**  
$$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin (x)} = 2 $$

---

### 4. $$ \lim _{x \rightarrow -2} \frac{x^{2}+5 x+6}{x-2} $$

**Evaluation:**

As $$ x \rightarrow -2 $$,
- Numerator $$ x^2 + 5x + 6 = (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 $$
- Denominator $$ x - 2 = -2 - 2 = -4 $$

The limit simplifies to:

$$
\frac{0}{-4} = 0
$$

**Result:**  
$$ \lim _{x \rightarrow -2} \frac{x^{2}+5 x+6}{x-2} = 0 $$

*Note:* L'Hôpital's Rule is **not necessary** here since the denominator does not approach zero.

---

### 5. $$ \lim _{x \rightarrow 2} \frac{x^{2}+5 x+6}{x-2} $$

**Evaluation:**

As $$ x \rightarrow 2 $$,
- Numerator $$ x^2 + 5x + 6 = 4 + 10 + 6 = 20 $$
- Denominator $$ x - 2 = 0 $$

This results in an infinite limit. To determine the direction:

- **From the right ($$ x \to 2^+ $$):**
  $$
  \frac{20}{\text{small positive}} \rightarrow +\infty
  $$

- **From the left ($$ x \to 2^- $$):**
  $$
  \frac{20}{\text{small negative}} \rightarrow -\infty
  $$

Since the left-hand limit and right-hand limit are not equal, the overall limit does **not exist** (it diverges to infinity).

**Result:**  
$$ \lim _{x \rightarrow 2} \frac{x^{2}+5 x+6}{x-2} $$ **does not exist** (it tends to $$ \pm\infty $$)
1. $$ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $$ 2. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} = \frac{1}{2} $$ 3. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin (x)} = 2 $$ 4. $$ \lim _{x \rightarrow -2} \frac{x^{2}+5 x+6}{x-2} = 0 $$ 5. $$ \lim _{x \rightarrow 2} \frac{x^{2}+5 x+6}{x-2} $$ **does not exist**

you apply the Rule of L’Hopital. Enter the letter “D” if the
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