Kalkulus Contoh

$lim_{x \to \infty} \frac{\sin^{2} (x)}{x^{2} + 1}$

Langkah 1

Karena $\frac{0}{x^{2} + 1} \leq \frac{\sin^{2} (x)}{x^{2} + 1} \leq \frac{1}{x^{2} + 1}$ dan $lim_{x \to \infty} \frac{0}{x^{2} + 1} = lim_{x \to \infty} \frac{1}{x^{2} + 1} = 0$ , terapkan Teorema Apit.

$0$