What is the value of `lim_(x to 0) (xsin5x)/(sin^(2)4x)`

Options

A. 0
B. `(5)/(4)`
C. `(5)/(16)`
D. `(25)/(4)`

Explanation

To find the limit \( \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} \]
To simplify the expression, we can multiply and divide by \( 5 \) and \( 16 \) to manipulate the sine functions:
\[ = \lim_{x \to 0} \frac{x \sin(5x)}{5x} \cdot \frac{5}{\sin^2(4x)} \cdot \frac{16(4x)^2}{16(4x)^2} \]

### Step 2: Separate the Limit
Now we can separate the limit into two parts:
\[ = \lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \cdot \frac{5}{\sin^2(4x)} \cdot \frac{16(4x)^2}{16(4x)^2} \right) \]
This gives us:
\[ = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot 5 \cdot \frac{16(4x)^2}{\sin^2(4x)} \]

### Step 3: Evaluate Each Limit
As \( x \to 0 \):
1. \( \lim_{x \to 0} \frac{\sin(5x)}{5x} = 1 \)
2. For the second part, we can rewrite \( \frac{16(4x)^2}{\sin^2(4x)} \) as \( \frac{16 \cdot 16x^2}{\sin^2(4x)} \):
\[ = \frac{16 \cdot 16x^2}{(4x)^2} \cdot \frac{(4x)^2}{\sin^2(4x)} \]
Hence,
\[ \lim_{x \to 0} \frac{(4x)^2}{\sin^2(4x)} = 1 \]

### Step 4: Combine the Results
Now we can combine the results:
\[ = 1 \cdot 5 \cdot 1 = 5 \]

### Step 5: Final Calculation
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} = \frac{5}{16} \]

### Final Answer
The value of the limit is:
\[ \frac{5}{16} \]

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