`lim_(x to 0) ((1- cos 2x)sin 5x)/(x^(2) sin 3x)`=

Options

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

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 expression using the cosine identity
We know that \( 1 - \cos 2x = 2 \sin^2 x \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{(2 \sin^2 x) \sin 5x}{x^2 \sin 3x} \]

### Step 2: Simplify the expression
Now we can simplify the limit:
\[ = \lim_{x \to 0} \frac{2 \sin^2 x \sin 5x}{x^2 \sin 3x} \]

### Step 3: Split the limit
We can separate the limit into parts:
\[ = 2 \lim_{x \to 0} \frac{\sin^2 x}{x^2} \cdot \lim_{x \to 0} \frac{\sin 5x}{\sin 3x} \]

### Step 4: Evaluate the first limit
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we have:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2 = 1^2 = 1 \]

### Step 5: Evaluate the second limit
For the second limit, 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 6: Combine the results
Now we can combine the results:
\[ = 2 \cdot 1 \cdot \frac{5}{3} = \frac{10}{3} \]

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

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