I don't quite understand this after diving the numerator by x² aren't we supposed to be left with 3/x

When we divide the entire numerator by $x^2$, we divide each term in the numerator separately:

$$\frac{2 + 5x + x \sin x}{x^2} = \frac{2}{x^2} + \frac{5x}{x^2} + \frac{x \sin x}{x^2}$$

This simplifies to:

$$\frac{2}{x^2} + \frac{5}{x} + \frac{\sin x}{x}$$

As $x \to \infty$:

• $\frac{2}{x^2} \to 0$
• $\frac{5}{x} \to 0$
• $\frac{\sin x}{x} \to 0$ (due to the Squeeze Theorem, as $-1 \le \sin x \le 1$)

So, the entire numerator approaches $0 + 0 + 0 = 0$.

The term $\frac{3}{x}$ actually appears in the denominator after dividing by $x^2$:

$$\frac{x^2 - 3x}{x^2} = \frac{x^2}{x^2} - \frac{3x}{x^2} = 1 - \frac{3}{x}$$

As $x \to \infty$, this denominator approaches $1 - 0 = 1$.

Therefore, the limit is $\frac{0}{1} = 0$.