`lim_(x -0) (1 - cos 4x)/(x^(2))` is equal to

Options

A. 0
B. 2
C. 4
D. 8

Answer

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos(4x)}{x^2} \), we can follow these steps:

### Step 1: Recognize the standard limit
We know from calculus that:
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \]
We will use this standard limit to help us evaluate our limit.

### Step 2: Rewrite the limit
We can rewrite our limit in a form that resembles the standard limit. Notice that we have \( \cos(4x) \) instead of \( \cos(x) \). To make it fit the standard form, we can manipulate the expression:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{x^2} = \lim_{x \to 0} \frac{1 - \cos(4x)}{(4x)^2} \cdot \frac{(4x)^2}{x^2} \]
This allows us to introduce \( (4x)^2 \) in the denominator.

### Step 3: Simplify the expression
Now, we can simplify the expression:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{(4x)^2} \cdot \frac{(4x)^2}{x^2} = \lim_{x \to 0} \frac{1 - \cos(4x)}{(4x)^2} \cdot 16 \]
Here, we have multiplied and divided by \( 16 \) because \( (4x)^2 = 16x^2 \).

### Step 4: Apply the standard limit
Now we can apply the standard limit we recognized earlier:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{(4x)^2} = \frac{1}{2} \]
Thus, we have:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{x^2} = 16 \cdot \frac{1}{2} = 8 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{1 - \cos(4x)}{x^2} = 8 \]