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Let's go through the limits for exercises 35 through 38 step by step.

### **Exercise 35: \(\lim_{x \to 0} \frac{1 - \cos x}{x}\)**

1. **Substitute \(x = 0\)** into the function:
   \[
   \frac{1 - \cos 0}{0} = \frac{1 - 1}{0} = \frac{0}{0}
   \]
   This results in an indeterminate form \(\frac{0}{0}\), so we use L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule:** Differentiate the numerator and denominator separately:
   \[
   \lim_{x \to 0} \frac{-\sin x}{1} = \lim_{x \to 0} (-\sin x)
   \]

3. **Substitute \(x = 0\)** again:
   \[
   -\sin 0 = 0
   \]
   The limit is **0**.

### **Exercise 36: \(\lim_{x \to 1} \frac{x^6 - 1}{(x - 1)^{1/3}}\)**

1. **Substitute \(x = 1\)** into the function:
   \[
   \frac{1^6 - 1}{(1 - 1)^{1/3}} = \frac{0}{0}
   \]
   This is an indeterminate form, so we apply L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule:** Differentiate the numerator and the denominator:
   \[
   \text{Numerator: } \frac{d}{dx}(x^6 - 1) = 6x^5
   \]
   \[
   \text{Denominator: } \frac{d}{dx}((x - 1)^{1/3}) = \frac{1}{3}(x - 1)^{-2/3}
   \]

3. **Rewrite the limit:**
   \[
   \lim_{x \to 1} \frac{6x^5}{\frac{1}{3}(x - 1)^{-2/3}} = \lim_{x \to 1} 18x^5(x - 1)^{2/3}
   \]

4. **Substitute \(x = 1\):**
   \[
   18 \cdot 1^5 \cdot (1 - 1)^{2/3} = 18 \times 0 = 0
   \]
   The limit is **0**.

### **Exercise 37: \(\lim_{x \to 0} \frac{\sin 2x}{x}\)**

1. **Direct Substitution:**


Explore exercises 35-38 on limits with detailed step-by-step solutions. Covering topics like L'Hôpital's Rule and trigonometric functions, these exercises provide a thorough understanding of calculus concepts. Ideal for advanced high school or college students.

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