Evaluate the limit _x infinity (x2 - x)/(x3 + 2x) using L'Hôpital's rule and differentiate y = 32.

1. Evaluate $\lim_{x \to \infty} \frac{x^2 - x}{x^3 + 2x}$ using L'Hôpital's rule.

Step 1: Check the form of the limit.
As $x \to \infty$, the numerator $x^2 - x \to \infty$ and the denominator $x^3 + 2x \to \infty$.
This is an indeterminate form $\frac{\infty}{\infty}$, so L'Hôpital's rule can be applied.

Step 2: Apply L'Hôpital's rule by differentiating the numerator and the denominator.

$$\lim_{x \to \infty} \frac{\frac{d}{dx}(x^2 - x)}{\frac{d}{dx}(x^3 + 2x)} = \lim_{x \to \infty} \frac{2x - 1}{3x^2 + 2}$$

Step 3: Check the form of the new limit.
As $x \to \infty$, the numerator $2x - 1 \to \infty$ and the denominator $3x^2 + 2 \to \infty$.
This is still an indeterminate form $\frac{\infty}{\infty}$, so apply L'Hôpital's rule again.

Step 4: Apply L'Hôpital's rule a second time.

$$\lim_{x \to \infty} \frac{\frac{d}{dx}(2x - 1)}{\frac{d}{dx}(3x^2 + 2)} = \lim_{x \to \infty} \frac{2}{6x}$$

Step 5: Evaluate the limit.

$$\lim_{x \to \infty} \frac{2}{6x} = \lim_{x \to \infty} \frac{1}{3x} = 0$$

The limit is $\boxed{0}$.

2. Differentiate the following functions: