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The image contains an example asking to show two limits using a known trigonometric limit:

1. **Using** \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \)
2. **Show that:**
    \[
    \lim_{h \to 0} \frac{\cos h - 1}{h} = 0
    \]

### Solution for the first limit:

We need to evaluate the limit:
\[
\lim_{h \to 0} \frac{\cos h - 1}{h}
\]

#### Step-by-Step Approach:

1. **Taylor Series Expansion of \(\cos h\)**:
   The Taylor series expansion of \(\cos h\) around \(h = 0\) is:
   \[
   \cos h = 1 - \frac{h^2}{2} + \frac{h^4}{24} + O(h^6)
   \]
   
2. **Substitute this expansion into the expression**:
   \[
   \frac{\cos h - 1}{h} = \frac{\left(1 - \frac{h^2}{2} + O(h^4)\right) - 1}{h} = \frac{- \frac{h^2}{2} + O(h^4)}{h}
   \]
   
3. **Simplify the expression**:
   \[
   \frac{- \frac{h^2}{2} + O(h^4)}{h} = -\frac{h}{2} + O(h^3)
   \]
   
4. **Take the limit as \(h \to 0\)**:
   As \(h \to 0\), both \(-\frac{h}{2}\) and the higher-order terms \(O(h^3)\) tend to zero. Therefore, the limit is:
   \[
   \lim_{h \to 0} \frac{\cos h - 1}{h} = 0
   \]

This proves that the limit is indeed 0.

Would you like further details or explanations on this?

Here are 5 related questions that expand upon the problem:

1. How is the Taylor series for other trigonometric functions derived?
2. What are the higher-order terms in the Taylor series for \(\sin h\)?
3. Why does the small-angle approximation work for trigonometric functions near zero?
4. How can L'Hopital's Rule be used to solve this limit problem?
5. Can you find the limit of \( \lim_{h \to 0} \frac{\sin h - h}{h^3} \)?

**Tip**: When dealing with trigonometric limits, Taylor series expansions around \(h = 0\) often provide a clearer path to the solution.

Show that (a) \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \)

Learn how to prove the limit \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \) by utilizing the known trigonometric limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) and the Taylor series expansion for cos(h). Suitable for students studying calculus or advanced mathematics.

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