Soal
Jika \(f(x) = \sin{2x}\), maka $$ \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + f(x-\frac{h}{2}) } {h^2} = \ldots $$
a. \(2\sin{2x}\)
b. \(\sin{2x}\)
c. 0
d. \(-\sin{2x}\)
e. \(2\sin{2x}\)
Jawab
Perhatikan bahwa jika, kita subtitusi nilai \(h = 0\) ke limit, maka diperoleh bentuk \(\frac{0}{0}\), sehingga nilai limit kita cari menggunakan turunan terhadap variabel \(h\).
$$
\begin{align} \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + (x-\frac{h}{2}) } {h^2} &= \lim_{h \to 0} \frac{\frac{1}{2}f'(x+\frac{h}{2}) + 0 + \left( -\frac{1}{2} \right)f'(x-\frac{h}{2}) } {2h}\\ &= \lim_{h \to 0} \frac{\frac{1}{4}f”(x+\frac{h}{2}) + \left( \frac{1}{4} \right)f”(x-\frac{h}{2}) } {2}\\ &= \frac{\frac{1}{4}f”(x+\frac{0}{2}) + \frac{1}{4} f”(x-\frac{0}{2}) } {2}\\ &= \frac{\frac{1}{4}f”(x) + \frac{1}{4} f”(x) } {2}\\ &= \frac{1}{4} f”(x) \end{align} $$
Karena \(f(x) = \sin{2x}\), maka \(f'(x) = 2 \cos{2x}\) dan \(f”(x) = -4 \sin{2x}\). Dengan demikian, $$ \begin{align} \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + (x-\frac{h}{2}) } {h^2} &= \frac{1}{4} f”(x) \\ &= \frac{1}{4} \cdot -4 \sin{2x} = -\sin{2x} \end{align} $$
