Evaluate: `lim_(xto oo)(2x^(2)-2x-sin^(2)x)/(3x^(2)-4x+cos^(2)x)`

Options

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

Answer

To evaluate the limit

\[ \lim_{x \to \infty} \frac{2x^2 - 2x - \sin^2 x}{3x^2 - 4x + \cos^2 x}, \]

we start by substituting \(x\) with \(\infty\). This gives us an indeterminate form of \(\frac{\infty}{\infty}\). To resolve this, we can factor out the highest power of \(x\) from both the numerator and the denominator.

### Step 1: Factor out \(x^2\)

We can rewrite the limit as follows:

\[ \lim_{x \to \infty} \frac{x^2(2 - \frac{2}{x} - \frac{\sin^2 x}{x^2})}{x^2(3 - \frac{4}{x} + \frac{\cos^2 x}{x^2})}. \]

### Step 2: Simplify the expression

Now, we can cancel \(x^2\) from the numerator and the denominator:

\[ \lim_{x \to \infty} \frac{2 - \frac{2}{x} - \frac{\sin^2 x}{x^2}}{3 - \frac{4}{x} + \frac{\cos^2 x}{x^2}}. \]

### Step 3: Evaluate the limit

As \(x\) approaches infinity, \(\frac{2}{x}\) and \(\frac{4}{x}\) both approach \(0\), and \(\frac{\sin^2 x}{x^2}\) and \(\frac{\cos^2 x}{x^2}\) also approach \(0\) since \(\sin^2 x\) and \(\cos^2 x\) are bounded between \(0\) and \(1\).

Thus, we can simplify the limit to:

\[ \frac{2 - 0 - 0}{3 - 0 + 0} = \frac{2}{3}. \]

### Final Answer

Therefore, the limit is:

\[ \lim_{x \to \infty} \frac{2x^2 - 2x - \sin^2 x}{3x^2 - 4x + \cos^2 x} = \frac{2}{3}. \]

---