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 expression:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} \]
We can rewrite this as:
\[ \lim_{x \to 0} \left( \frac{\sin(4x)}{x} \right)^2 \]

### Step 2: Adjust the expression
To apply the standard limit \( \lim_{y \to 0} \frac{\sin(y)}{y} = 1 \), we need to manipulate the expression so that the argument of the sine function matches the denominator. We can do this by multiplying and dividing by 4:
\[ \lim_{x \to 0} \left( \frac{\sin(4x)}{4x} \cdot \frac{4x}{x} \right)^2 \]
This simplifies to:
\[ \lim_{x \to 0} \left( \frac{\sin(4x)}{4x} \cdot 4 \right)^2 \]

### Step 3: Apply the limit
Now we can separate the limit:
\[ = \left( \lim_{x \to 0} \frac{\sin(4x)}{4x} \cdot 4 \right)^2 \]
As \( x \to 0 \), \( \frac{\sin(4x)}{4x} \) approaches 1. Therefore:
\[ = (1 \cdot 4)^2 \]

### Step 4: Calculate the final result
This simplifies to:
\[ = 4^2 = 16 \]

Thus, the final result is:
\[ \lim_{x \to 0} \frac{\sin^2(4x)}{x^2} = 16 \]