Evaluate `lim_(x to 0) ("sin"^(2) 4x)/(x^(2))`

Answer

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

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} \]

### Step 2: Introduce a factor to facilitate the limit
We can rewrite the expression by introducing \( \frac{4}{4} \) (which is equal to 1) in the limit:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} = \lim_{x \to 0} \frac{\sin^2(4x)}{(4x)^2} \cdot \frac{(4x)^2}{x^2} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{(4x)^2} \cdot 16 \]

### Step 3: Apply the limit property
Now we can apply the known limit property \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). Here, let \( u = 4x \). As \( x \to 0 \), \( u \to 0 \) as well. Thus:
\[ \lim_{x \to 0} \frac{\sin(4x)}{4x} = 1 \]
So:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{(4x)^2} = \left(\lim_{x \to 0} \frac{\sin(4x)}{4x}\right)^2 = 1^2 = 1 \]

### Step 4: Combine results
Now substituting this back into our expression:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} = 1 \cdot 16 = 16 \]

### Final Answer
Thus, we conclude that:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} = 16 \]