But what happened to the 1 ,after dividing x²/x² in the numerator aren't we supposed to get 1

In Step 3, the original expression was $\lim_{x \to \infty} \frac{2 + 5x + x \sin x}{x^2 - 3x}$.

We divided every term in both the numerator and the denominator by $x^2$:

• Numerator:

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

  As $x \to \infty$, each of these terms goes to 0. So, the numerator approaches $0+0+0 = 0$.

• Denominator:

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

  As $x \to \infty$, $\frac{x^2}{x^2}$ becomes 1, and $\frac{3}{x}$ goes to 0. So, the denominator approaches $1-0 = 1$.

Therefore, the limit becomes $\frac{0}{1} = 0$. The 1 from $\frac{x^2}{x^2}$ is indeed in the denominator, as shown in the calculation.