Lanjutan 2 Limit Fungsi Trigonometri

 A. Teorema Apit

Misalkan  $f$,  $g$, dan   $h$  adalah fungsi yang memenuhi  $f(x)\le g(x)\le h(x)$ untuk seluruh titik di sekitar  $c$.  

Jika  $\underset{x\to c}{\text{lim}}f(x)=\underset{x\to c}{\text{lim}}g(x)=\text{L}$,  maka   $\underset{x\to c}{\text{lim}}g(x)=\text{L}$.

B. Penentuan nilai  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}$  dan  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}}$.

Untuk bukti dari 

$\begin{array}{rl}1.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\\ 2.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\end{array}$.

Berikut penjabaran buktinya

Perhatikan ilustrasi gambar berikut ini

$\begin{array}{|cll|}\hline \text{No}& \text{Nama}& \text{Luas Bangun}\\ 1& \mathrm{△}\text{AOC}& \begin{array}{rl}{\text{L}}_{{}_{\mathrm{△}\text{AOC}}}& ={\displaystyle \frac{1}{2}.OA.CD}\\ & ={\displaystyle \frac{1}{2}.OA.OC.\sin x}\\ & ={\displaystyle \frac{1}{2}.r.r.\sin x}\\ & ={\displaystyle \frac{1}{2}{r}^2\sin x}\end{array}\\ 2& \text{Juring AOC}& \begin{array}{rl}{\text{L}}_{{}_{\text{Juring AOC}}}& ={\displaystyle \frac{x}{2\pi }.\pi r^2}\\ & ={\displaystyle \frac{x}{2\pi }.\pi .r^2}\\ & ={\displaystyle \frac{1}{2}x.r^2}\end{array}\\ 3& \mathrm{△}\text{AOB}& \begin{array}{rl}{\text{L}}_{{}_{\mathrm{△}\text{AOB}}}& ={\displaystyle \frac{1}{2}.OA.AB}\\ & ={\displaystyle \frac{1}{2}.OA.OA.\tan x}\\ & ={\displaystyle \frac{1}{2}.r.r.\tan x}\\ & ={\displaystyle \frac{1}{2}{r}^2\tan x}\end{array}\\ \hline\end{array}$.

Selanjutnya perhatikan pula bahwa dari fakta di atas dapat dituliskan sebagai berikut, yaitu:

$\begin{array}{|cc|}\hline \mathbf{\text{Bagian Pertama}}& \text{Bagian Kedua}\\ \begin{array}{r}\\ {\displaystyle \frac{1}{2}{r}^2\sin x}& \le & {\displaystyle \frac{1}{2}x{r}^2}& \le & {\displaystyle \frac{1}{2}{r}^2\tan x}\\ \\ \sin x& \le & x& \le & {\displaystyle \frac{\sin x}{\cos x}}\\ \\ 1& \le & {\displaystyle \frac{x}{\sin x}}& \le & {\displaystyle \frac{1}{\cos x}}\\ \\ 1& \ge & {\displaystyle \frac{\sin x}{x}}& \ge & \cos x\\ \\ \cos x& \le & {\displaystyle \frac{\sin x}{x}}& \le & 1\\ \\ & & & & \end{array}& \begin{array}{r}\\ {\displaystyle \frac{1}{2}{r}^2\sin x}& \le & {\displaystyle \frac{1}{2}x{r}^2}& \le & {\displaystyle \frac{1}{2}{r}^2\tan x}\\ \\ {\displaystyle \sin x}& \le & x& \le & {\displaystyle \tan x}\\ \\ {\displaystyle \frac{\sin x}{\tan x}}& \le & {\displaystyle \frac{x}{\tan x}}& \le & 1\\ \\ \cos x& \le & {\displaystyle \frac{x}{\tan x}}& \le & 1\\ \\ {\displaystyle \frac{1}{\cos x}}& \ge & {\displaystyle \frac{\tan x}{x}}& \ge & 1\\ 1& \le & {\displaystyle \frac{\tan x}{x}}& \le & {\displaystyle \frac{1}{\cos x}}\end{array}\\ \hline\end{array}$

Dan untuk mendapatkan nilai  yang diinginkan adalah:

$\begin{array}{rl}\mathbf{\text{Bag}}& \mathbf{\text{ian pertama}}:\\ & \underset{x\to 0}{\text{lim}}\cos x\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le \underset{x\to 0}{\text{lim}}1}\\ & \iff 1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le 1}\\ & \text{Dengan teorema apit},\text{maka}\\ & \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=1.}\\ \mathbf{\text{Bag}}& \mathbf{\text{ian kedua}}:\\ & \underset{x\to 0}{\text{lim}}1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}}}\\ & \iff 1\le \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\le 1}\\ & \text{Dengan teorema apit},\text{maka}\\ & \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=1.}\end{array}$.

Selanjutnya untuk mendapatkan nilai $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}}$ dan  $\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan x}}$ dapat diperoleh dari bagian pertama dan kedua di atas, yaitu:

$\begin{array}{rl}\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan x}{x}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\times \frac{1}{\cos x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}}}\\ & =1\times 1\\ & =1\\ \text{Demikia}& \text{n juga}\\ \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan x}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}\times \frac{x}{\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1}{\cos x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}}\\ & =1\times 1\\ & =1\end{array}$.

Dari uraian panjang di atas telah ditunjukkan dengan bukti-buktinya bahwa

$\begin{array}{rl}1.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\\ 2.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin x}{x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}=1}}\end{array}$

DAFTAR PUSTAKA

1. Kartini, Suprapto, Subandi, Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
2. 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.