Soal

Jika [latex]f(x) = \sin{2x}[/latex], maka $$ \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + f(x-\frac{h}{2}) } {h^2} = \ldots $$

a. [latex]2\sin{2x}[/latex]

b. [latex]\sin{2x}[/latex]

c. 0

d. [latex]-\sin{2x}[/latex]

e. [latex]2\sin{2x}[/latex]

Jawab

Perhatikan bahwa jika, kita subtitusi nilai [latex]h = 0[/latex] ke limit, maka diperoleh bentuk [latex]\frac{0}{0}[/latex], sehingga nilai limit kita cari menggunakan turunan terhadap variabel [latex]h[/latex].

$$
\begin{align} \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + (x-\frac{h}{2}) } {h^2} &= \lim_{h \to 0} \frac{\frac{1}{2}f'(x+\frac{h}{2}) + 0 + \left( -\frac{1}{2} \right)f'(x-\frac{h}{2}) } {2h}\\ &= \lim_{h \to 0} \frac{\frac{1}{4}f”(x+\frac{h}{2}) + \left( \frac{1}{4} \right)f”(x-\frac{h}{2}) } {2}\\ &= \frac{\frac{1}{4}f”(x+\frac{0}{2}) + \frac{1}{4} f”(x-\frac{0}{2}) } {2}\\ &= \frac{\frac{1}{4}f”(x) + \frac{1}{4} f”(x) } {2}\\ &= \frac{1}{4} f”(x) \end{align} $$

Karena [latex]f(x) = \sin{2x}[/latex], maka [latex]f'(x) = 2 \cos{2x}[/latex] dan [latex]f”(x) = -4 \sin{2x}[/latex]. Dengan demikian, $$ \begin{align} \lim_{h \to 0} \frac{f(x+\frac{h}{2}) + 2f(x) + (x-\frac{h}{2}) } {h^2} &= \frac{1}{4} f”(x) \\ &= \frac{1}{4} \cdot -4 \sin{2x} = -\sin{2x} \end{align} $$