A. Teorema Apit
Misalkan $f$, $g$, dan $h$ adalah fungsi yang memenuhi $f(x)\leq g(x)\leq h(x)$ untuk seluruh titik di sekitar $c$.
Jika $\underset{x\rightarrow c }{\textrm{lim}}\: f(x) =\underset{x\rightarrow c }{\textrm{lim}}\: g(x)=\textrm{L}$, maka $\underset{x\rightarrow c }{\textrm{lim}}\: g(x)=\textrm{L}$.
B. Penentuan nilai $\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}$ dan $\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}$.
Untuk bukti dari
$\color{red}\begin{aligned}\color{blue}1.\quad& \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=\color{black}1\\ \color{blue}2.\quad& \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=\color{black}1 \end{aligned}$.
Berikut penjabaran buktinya
Perhatikan ilustrasi gambar berikut ini
$\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Nama}&\qquad\textrm{Luas Bangun}\\\hline 1&\triangle \textrm{AOC}&\begin{aligned}\textrm{L}_{_{\triangle \textrm{AOC}}}&=\displaystyle \frac{1}{2}.OA.CD\\ &=\displaystyle \frac{1}{2}.OA.OC.\sin x\\ &=\displaystyle \frac{1}{2}.r.r.\sin x\\ &=\color{blue}\displaystyle \frac{1}{2}r^{2}\sin x \end{aligned}\\\hline 2&\textrm{Juring AOC}&\begin{aligned}\textrm{L}_{_{\textrm{Juring AOC}}}&=\displaystyle \frac{x}{2\pi }.\pi r^{2}\\ &=\displaystyle \frac{x}{2\pi }.\pi .r^{2}\\ &=\color{blue}\displaystyle \frac{1}{2}x.r^{2} \end{aligned}\\\hline 3&\triangle \textrm{AOB}&\begin{aligned}\textrm{L}_{_{\triangle \textrm{AOB}}}&=\displaystyle \frac{1}{2}.OA.AB\\ &=\displaystyle \frac{1}{2}.OA.OA.\tan x\\ &=\displaystyle \frac{1}{2}.r.r.\tan x\\ &=\color{blue}\displaystyle \frac{1}{2}r^{2}\tan x \end{aligned}\\\hline \end{array}$.
Selanjutnya perhatikan pula bahwa dari fakta di atas dapat dituliskan sebagai berikut, yaitu:
$\begin{array}{|c|c|}\hline \textbf{Bagian Pertama}&\textrm{Bagian Kedua}\\\hline \begin{array}{rlclll}\\ \displaystyle \frac{1}{2}r^{2}\sin x&\leq &\displaystyle \frac{1}{2}xr^{2}&\leq &\displaystyle \frac{1}{2}r^{2}\tan x\\\\ \sin x&\leq &x&\leq &\displaystyle \frac{\sin x}{\cos x}\\\\ 1&\leq &\displaystyle \frac{x}{\sin x}&\leq &\displaystyle \frac{1}{\cos x}\\\\ 1&\geq &\displaystyle \frac{\sin x}{x}&\geq &\cos x\\\\ \cos x&\leq &\displaystyle \frac{\sin x}{x}&\leq &1\\\\ &&&& \end{array}&\begin{array}{rlclll}\\ \displaystyle \frac{1}{2}r^{2}\sin x&\leq &\displaystyle \frac{1}{2}xr^{2}&\leq &\displaystyle \frac{1}{2}r^{2}\tan x\\\\ \displaystyle \sin x&\leq &x&\leq &\displaystyle \tan x\\\\ \displaystyle \frac{\sin x}{\tan x}&\leq &\displaystyle \frac{x}{\tan x}&\leq &1\\\\ \cos x&\leq &\displaystyle \frac{ x}{\tan x}&\leq &1\\\\ \displaystyle \frac{1}{\cos x}&\geq &\displaystyle \frac{\tan x}{x}&\geq &1\\ 1&\leq &\displaystyle \frac{\tan x}{x}&\leq &\displaystyle \frac{1}{\cos x} \end{array} \\\hline \end{array}$
Dan untuk mendapatkan nilai yang diinginkan adalah:
$\begin{aligned}\textbf{Bag}&\textbf{ian pertama}:\\ &\underset{x\rightarrow 0 }{\textrm{lim}}\: \cos x\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: 1\\ &\Leftrightarrow 1\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}\leq 1\\ &\color{red}\textrm{Dengan teorema apit},\: \color{black}\textrm{maka}\\ &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=1.\\ \textbf{Bag}&\textbf{ian kedua}:\\ &\underset{x\rightarrow 0 }{\textrm{lim}}\: 1\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan x}{x}\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1}{\cos x}\\ &\Leftrightarrow 1\leq \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}\leq 1\\ &\color{red}\textrm{Dengan teorema apit},\: \color{black}\textrm{maka}\\ &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=1.\\ \end{aligned}$.
Selanjutnya untuk mendapatkan nilai $\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan x}{x}$ dan $\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\tan x}$ dapat diperoleh dari bagian pertama dan kedua di atas, yaitu:
$\color{red}\begin{aligned} \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan x}{x}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}\color{black}\times \frac{1}{\cos x}\\ &\color{black}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1}{\cos x}\\ &\color{black}=1\times 1\\ &\color{black}=1\\ \color{black}\textrm{Demikia}& \color{black}\textrm{n juga}\\ \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\tan x}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1}{\cos x}\color{black}\times \frac{x}{\sin x}\\ &\color{black}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1}{\cos x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}\\ &\color{black}=1\times 1\\ &\color{black}=1\\ \end{aligned}$.
Dari uraian panjang di atas telah ditunjukkan dengan bukti-buktinya bahwa
$\begin{aligned}\color{blue}1.\quad& \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=\color{black}1\\ \color{blue}2.\quad& \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=\color{black}1 \end{aligned}$
DAFTAR PUSTAKA
- Kartini, Suprapto, Subandi, Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
- Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
