Misalkan fungsi f memenuhi |x^2 f(x) + 1| <= sin^2 (x-2) utk x real. Dengan Teorema Apit, tentukan lim x->2 f(x).

Materi Limit

[tex] \displaystyle |x^2f(x)+1| \leq \sin^2(x-2) \\ -\sin^2(x-2) \leq x^2f(x)+1 \leq \sin^2(x-2) \\ \frac{-\sin^2(x-2) - 1}{x^2} \leq f(x) \leq \frac{\sin^2(x-2)-1}{x^2} \\ \text{Perhatikan bahwa} \\ \lim_{x \to 2} \frac{-\sin^2(x-2) - 1}{x^2} = -\frac{1}{4} = \lim_{x \to 2} \frac{\sin^2(x-2) - 1}{x^2} \\ \text{Dengan Teorema Apit, dapat disimpulkan bahwa} \\ \lim_{x \to 2} f(x) = -\frac{1}{4} [/tex]