Exercise: 1. Evaluate (a) $$ \lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 e^{x}-2-x^{2}} $$ (b) $$ \lim _{x \rightarrow \infty} \frac{7 x-28}{x^{3}} $$ (c) $$ \lim _{x \rightarrow 1} \frac{x^{2}-x-2}{x^{2}-1} $$ 2. Evaluate (a) $$ \lim _{x \rightarrow 1} x^{\frac{1}{x-1}} $$. (b) $$ \lim _{x \rightarrow 0} x^{\sin x} $$ (c) $$ \lim _{x \rightarrow 0} x^{x} $$.

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UpStudy AI · Accepted Answer

Sure, let's evaluate each of the given limits step by step.

---

### **1. Evaluations**

#### **1(a)**
$$
\lim_{x \to 0} \frac{2 \sin x - \sin 2x}{2 e^{x} - 2 - x^{2}}
$$

**Approach: Taylor Series Expansion**

- **Numerator:**
  $$
  2 \sin x - \sin 2x = 2\left(x - \frac{x^3}{6}\right) - \left(2x - \frac{8x^3}{6}\right) + \text{higher terms} = \frac{x^3}{2} + \text{higher terms}
  $$

- **Denominator:**
  $$
  2 e^x - 2 - x^2 = 2\left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) - 2 - x^2 = 2x + \frac{x^3}{3} + \text{higher terms}
  $$

- **Limit:**
  $$
  \lim_{x \to 0} \frac{\frac{x^3}{2}}{2x + \frac{x^3}{3}} = \lim_{x \to 0} \frac{x^2}{4} = 0
  $$

**Alternatively, using L’Hôpital’s Rule:**

Both numerator and denominator approach 0 as $$ x \to 0 $$. Differentiating:

- **First Differentiation:**
  $$
  \frac{d}{dx}(2 \sin x - \sin 2x) = 2\cos x - 2\cos 2x
  $$
  $$
  \frac{d}{dx}(2 e^{x} - 2 - x^{2}) = 2 e^{x} - 2x
  $$
  
  Evaluating at $$ x = 0 $$:
  $$
  \frac{2(1) - 2(1)}{2(1) - 0} = \frac{0}{2} = 0
  $$

**Final Answer:**
$$
\boxed{0}
$$

---

#### **1(b)**
$$
\lim_{x \to \infty} \frac{7x - 28}{x^{3}}
$$

**Approach: Dominant Terms**

As $$ x \to \infty $$, the dominant term in the numerator is $$ 7x $$ and in the denominator is $$ x^3 $$.

$$
\lim_{x \to \infty} \frac{7x}{x^{3}} = \lim_{x \to \infty} \frac{7}{x^{2}} = 0
$$

**Final Answer:**
$$
\boxed{0}
$$

---

#### **1(c)**
$$
\lim_{x \to 1} \frac{x^{2} - x - 2}{x^{2} - 1}
$$

**Approach: Factorization**

- **Factor Numerator and Denominator:**
  $$
  \frac{x^{2} - x - 2}{x^{2} - 1} = \frac{(x - 2)(x + 1)}{(x - 1)(x + 1)} = \frac{x - 2}{x - 1}, \quad x \neq -1
  $$

- **Evaluate the Limit:**
  $$
  \lim_{x \to 1} \frac{x - 2}{x - 1} = \frac{-1}{0}
  $$
  
  - **From the Right ($$ x \to 1^+ $$):**
    $$
    \frac{-1}{0^+} = -\infty
    $$
  
  - **From the Left ($$ x \to 1^- $$):**
    $$
    \frac{-1}{0^-} = +\infty
    $$
  
  Since the one-sided limits are not equal, the two-sided limit does not exist.

**Final Answer:**
$$
\boxed{\text{The limit does not exist; it approaches } -\infty \text{ from above and } +\infty \text{ from below.}}
$$

---

### **2. Evaluations**

#### **2(a)**
$$
\lim_{x \to 1} x^{\frac{1}{x-1}}
$$

**Approach: Logarithmic Transformation**

Let $$ L = \lim_{x \to 1} x^{\frac{1}{x-1}} $$. Taking natural logarithm:

$$
\ln L = \lim_{x \to 1} \frac{\ln x}{x - 1}
$$

Using the approximation $$ \ln x \approx (x - 1) $$ as $$ x \to 1 $$:

$$
\ln L = \lim_{x \to 1} \frac{x - 1}{x - 1} = 1 \implies L = e^1 = e
$$

**Final Answer:**
$$
\boxed{e}
$$

---

#### **2(b)**
$$
\lim_{x \to 0} x^{\sin x}
$$

**Approach: Logarithmic Transformation**

Let $$ L = \lim_{x \to 0} x^{\sin x} $$. Taking natural logarithm:

$$
\ln L = \lim_{x \to 0} \sin x \cdot \ln x
$$

As $$ x \to 0^+ $$, $$ \sin x \approx x $$ and $$ \ln x \to -\infty $$. Thus:

$$
\ln L = \lim_{x \to 0^+} x \cdot (-\infty) = 0
$$

$$
L = e^0 = 1
$$

**Final Answer:**
$$
\boxed{1}
$$

---

#### **2(c)**
$$
\lim_{x \to 0} x^{x}
$$

**Approach: Logarithmic Transformation**

Let $$ L = \lim_{x \to 0^+} x^{x} $$. Taking natural logarithm:

$$
\ln L = \lim_{x \to 0^+} x \cdot \ln x
$$

As $$ x \to 0^+ $$, $$ \ln x \to -\infty $$, but $$ x \cdot \ln x \to 0 $$:

$$
\ln L = 0 \implies L = e^0 = 1
$$

**Final Answer:**
$$
\boxed{1}
$$

---

### **Summary of Answers**

1. **Evaluations**
   - **1(a):** $$ \boxed{0} $$
   - **1(b):** $$ \boxed{0} $$
   - **1(c):** $$ \boxed{\text{Does not exist; approaches } -\infty \text{ and } +\infty} $$

2. **Evaluations**
   - **2(a):** $$ \boxed{e} $$
   - **2(b):** $$ \boxed{1} $$
   - **2(c):** $$ \boxed{1} $$
**1. Evaluations** 1(a) The limit is 0. 1(b) The limit is 0. 1(c) The limit does not exist; it approaches negative infinity from one side and positive infinity from the other. --- **2. Evaluations** 2(a) The limit is $$ e $$. 2(b) The limit is 1. 2(c) The limit is 1. **Summary:** - 1(a): 0 - 1(b): 0 - 1(c): Does not exist - 2(a): $$ e $$ - 2(b): 1 - 2(c): 1

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