Evaluate the following limits :
`Lim_(x to 0 ) (1- cos ax)/(" x sin 3x")`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(ax)}{x \sin(3x)}, \]

we can use L'Hôpital's Rule, which is applicable when we encounter an indeterminate form like \( \frac{0}{0} \).

### Step-by-Step Solution:

1. **Check the form of the limit:**
As \( x \to 0 \):
- \( 1 - \cos(ax) \to 1 - 1 = 0 \)
- \( x \sin(3x) \to 0 \cdot 0 = 0 \)

Thus, we have the indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule:**
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- **Numerator:** The derivative of \( 1 - \cos(ax) \) is \( a \sin(ax) \).
- **Denominator:** The derivative of \( x \sin(3x) \) can be found using the product rule:
\[ \frac{d}{dx}(x \sin(3x)) = \sin(3x) + x \cdot 3 \cos(3x) = \sin(3x) + 3x \cos(3x). \]

Therefore, we can rewrite our limit as:
\[ \lim_{x \to 0} \frac{a \sin(ax)}{\sin(3x) + 3x \cos(3x)}. \]

3. **Evaluate the new limit:**
As \( x \to 0 \):
- \( \sin(ax) \to ax \)
- \( \sin(3x) \to 3x \)
- \( 3x \cos(3x) \to 3x \cdot 1 = 3x \)

Thus, we substitute these limits:
\[ \lim_{x \to 0} \frac{a(ax)}{3x + 3x} = \lim_{x \to 0} \frac{a^2 x}{6x} = \lim_{x \to 0} \frac{a^2}{6} = \frac{a^2}{6}. \]

### Final Answer:
\[ \lim_{x \to 0} \frac{1 - \cos(ax)}{x \sin(3x)} = \frac{a^2}{6}. \]