`lim_(x to 0) (x sin5x)/(sin^(2)4x)` equal to ?

Options

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

Answer

To solve 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 expression:
\[ \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} \]

### Step 2: Use the limit property of sine
We know that as \( x \to 0 \), \( \frac{\sin(kx)}{kx} \to 1 \) for any constant \( k \). We can rewrite the sine terms in the limit:
\[ \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} = \lim_{x \to 0} \frac{x \cdot \sin(5x)}{(4x)^2} \cdot \frac{(4x)^2}{\sin^2(4x)} \]

### Step 3: Substitute and simplify
Now, we can substitute:
\[ = \lim_{x \to 0} \frac{x \cdot \sin(5x)}{(4x)^2} \cdot \frac{(4x)^2}{\sin^2(4x)} = \lim_{x \to 0} \frac{x \cdot \sin(5x)}{16x^2} \cdot \frac{16x^2}{\sin^2(4x)} \]

### Step 4: Break it down further
This can be simplified to:
\[ = \lim_{x \to 0} \frac{\sin(5x)}{4x} \cdot \frac{16x^2}{\sin^2(4x)} \]

### Step 5: Apply the limit property
Using the property \( \frac{\sin(kx)}{kx} \to 1 \):
\[ = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5}{4} \cdot \frac{16x^2}{\sin^2(4x)} \]
\[ = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5}{4} \cdot \left(\frac{4}{\sin(4x)}\right)^2 \]

### Step 6: Evaluate the limits
As \( x \to 0 \):
\[ \frac{\sin(5x)}{5x} \to 1 \quad \text{and} \quad \left(\frac{4}{\sin(4x)}\right)^2 \to 1 \]
Thus, we have:
\[ = 1 \cdot \frac{5}{4} \cdot 1 = \frac{5}{4} \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{x \sin(5x)}{\sin^2(4x)} = \frac{5}{4} \]