`lim_(x to oo)(sin^(4)x-sin^(2)x+1)/(cos^(4)x-cos^(2)x+1)` is equal to

Options

A. 0
B. 1
C. `1//3`
D. `1//2`

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{\sin^4 x - \sin^2 x + 1}{\cos^4 x - \cos^2 x + 1}, \]

we can follow these steps:

### Step 1: Rewrite the limit expression

We start by rewriting the limit expression to make it easier to analyze:

\[ \lim_{x \to \infty} \frac{\sin^4 x - \sin^2 x + 1}{\cos^4 x - \cos^2 x + 1}. \]

### Step 2: Substitute trigonometric identities

Recall that \( \sin^2 x + \cos^2 x = 1 \). We can express \( \cos^2 x \) in terms of \( \sin^2 x \):

\[ \cos^2 x = 1 - \sin^2 x. \]

Now, we can rewrite the denominator:

\[ \cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x. \]

Substituting this into the limit gives:

\[ \lim_{x \to \infty} \frac{\sin^4 x - \sin^2 x + 1}{(1 - 2\sin^2 x + \sin^4 x) - (1 - \sin^2 x) + 1}. \]

### Step 3: Simplify the denominator

Now simplify the denominator:

\[ \cos^4 x - \cos^2 x + 1 = (1 - 2\sin^2 x + \sin^4 x) - (1 - \sin^2 x) + 1. \]

This simplifies to:

\[ -2\sin^2 x + \sin^4 x + \sin^2 x + 1 = \sin^4 x - \sin^2 x + 1. \]

### Step 4: Substitute back into the limit

Now we have:

\[ \lim_{x \to \infty} \frac{\sin^4 x - \sin^2 x + 1}{\sin^4 x - \sin^2 x + 1}. \]

### Step 5: Evaluate the limit

Since the numerator and denominator are the same, we can simplify this to:

\[ \lim_{x \to \infty} 1 = 1. \]

### Final Answer

Thus, the limit is:

\[ \boxed{1}. \]