`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`

Options

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

Explanation

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 start by rewriting the limit expression:
\[ \lim_{x \to 0} \frac{(1 - \cos 2x) \sin 5x}{x^2 \sin 3x} \]

### Step 2: Use known limits
We can use the following known limits:
1. \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \)
2. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

### Step 3: Rewrite \( 1 - \cos 2x \)
Using the first limit, we can express \( 1 - \cos 2x \) in terms of \( x^2 \):
\[ 1 - \cos 2x = 2 \sin^2(x) \]
Thus,
\[ \frac{1 - \cos 2x}{x^2} = \frac{2 \sin^2 x}{x^2} = 2 \left(\frac{\sin x}{x}\right)^2 \]

### Step 4: Substitute into the limit
Now we can rewrite our limit:
\[ \lim_{x \to 0} \frac{(1 - \cos 2x) \sin 5x}{x^2 \sin 3x} = \lim_{x \to 0} \frac{2 \left(\frac{\sin x}{x}\right)^2 \sin 5x}{\sin 3x} \]

### Step 5: Analyze \( \sin 5x \) and \( \sin 3x \)
Using the second limit:
\[ \lim_{x \to 0} \frac{\sin 5x}{5x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin 3x}{3x} = 1 \]
Thus, we can express:
\[ \sin 5x \approx 5x \quad \text{and} \quad \sin 3x \approx 3x \]

### Step 6: Substitute these approximations
Substituting these approximations back into the limit:
\[ \lim_{x \to 0} \frac{2 \left(\frac{\sin x}{x}\right)^2 \cdot 5x}{3x} \]

### Step 7: Simplify the expression
Now, we simplify the expression:
\[ = \lim_{x \to 0} \frac{2 \cdot 5}{3} \cdot \left(\frac{\sin x}{x}\right)^2 \]

### Step 8: Evaluate the limit
Since \( \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 1 \):
\[ = \frac{10}{3} \]

### Conclusion
Thus, the final result is:
\[ \lim_{x \to 0} \frac{(1 - \cos 2x) \sin 5x}{x^2 \sin 3x} = \frac{10}{3} \]