`Lt_(xto0)((1-cos2x)sin5x)/(x^(2)sin3x)` is equal to
(i) `(6)/(5)`
(ii) `5/6`
(iii) `10/3`
(iv) `3/10`

Options

A. `(6)/(5)`
B. `5/6`
C. `10/3`
D. `3/10`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(1 - \cos 2x) \sin 5x}{x^2 \sin 3x} \), we can follow these steps:

### Step 1: Rewrite the limit
We can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{1 - \cos 2x}{x^2} \cdot \lim_{x \to 0} \frac{\sin 5x}{\sin 3x} \]

### Step 2: Use the identity for \( 1 - \cos 2x \)
Using the trigonometric identity \( 1 - \cos 2x = 2 \sin^2 x \), we can rewrite the limit:
\[ \lim_{x \to 0} \frac{2 \sin^2 x}{x^2} \cdot \lim_{x \to 0} \frac{\sin 5x}{\sin 3x} \]

### Step 3: Simplify the first limit
The limit \( \lim_{x \to 0} \frac{\sin^2 x}{x^2} \) can be simplified using the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ \lim_{x \to 0} \frac{2 \sin^2 x}{x^2} = 2 \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2 = 2 \cdot 1^2 = 2 \]

### Step 4: Simplify the second limit
For the second limit \( \lim_{x \to 0} \frac{\sin 5x}{\sin 3x} \), we can use the fact that \( \lim_{x \to 0} \frac{\sin kx}{kx} = 1 \):
\[ \lim_{x \to 0} \frac{\sin 5x}{\sin 3x} = \lim_{x \to 0} \frac{5x}{3x} = \frac{5}{3} \]

### Step 5: Combine the limits
Now we can combine the results from the two limits:
\[ \lim_{x \to 0} \frac{(1 - \cos 2x) \sin 5x}{x^2 \sin 3x} = 2 \cdot \frac{5}{3} = \frac{10}{3} \]

### Final Answer
Thus, the limit is:
\[ \frac{10}{3} \]