Evaluate `lim_(x to 0) (1 - cos ax)/(1 - cos bx)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos(ax)}{1 - \cos(bx)} \), we start by substituting \( x = 0 \):

1. **Substitution**:
\[ \frac{1 - \cos(a \cdot 0)}{1 - \cos(b \cdot 0)} = \frac{1 - \cos(0)}{1 - \cos(0)} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \]
This gives us an indeterminate form \( \frac{0}{0} \).

**Hint**: When you encounter a \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

2. **Applying L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
Here, let \( f(x) = 1 - \cos(ax) \) and \( g(x) = 1 - \cos(bx) \).

3. **Finding Derivatives**:
- The derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(1 - \cos(ax)) = a \sin(ax) \]
- The derivative of \( g(x) \):
\[ g'(x) = \frac{d}{dx}(1 - \cos(bx)) = b \sin(bx) \]

4. **Reapplying the Limit**:
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{a \sin(ax)}{b \sin(bx)} \]

5. **Substituting \( x = 0 \) Again**:
Substituting \( x = 0 \):
\[ = \frac{a \sin(a \cdot 0)}{b \sin(b \cdot 0)} = \frac{a \cdot 0}{b \cdot 0} = \frac{0}{0} \]
This is still an indeterminate form, so we apply L'Hôpital's Rule again.

6. **Finding Second Derivatives**:
- The second derivative of \( f(x) \):
\[ f''(x) = \frac{d}{dx}(a \sin(ax)) = a^2 \cos(ax) \]
- The second derivative of \( g(x) \):
\[ g''(x) = \frac{d}{dx}(b \sin(bx)) = b^2 \cos(bx) \]

7. **Reapplying the Limit**:
Now we have:
\[ \lim_{x \to 0} \frac{a^2 \cos(ax)}{b^2 \cos(bx)} \]

8. **Substituting \( x = 0 \) Again**:
Substituting \( x = 0 \):
\[ = \frac{a^2 \cos(a \cdot 0)}{b^2 \cos(b \cdot 0)} = \frac{a^2 \cdot 1}{b^2 \cdot 1} = \frac{a^2}{b^2} \]

Thus, the final result is:
\[ \lim_{x \to 0} \frac{1 - \cos(ax)}{1 - \cos(bx)} = \frac{a^2}{b^2} \]

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