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Membuktikan Limit Barisan Menggunakan Definisi

Pengertian barisan yang dimaksud di sini adalah pengertian barisan bilangan real, yaitu suatu fungsi pada himpunan $N$ dengan daerah hasil yang termuat di $R$. Dengan kata lain, suatu barisan di $R$ memasangkan masing-masing bilangan asli 1, 2, 3, dst. secara tunggal dengan bilangan real. Bilangan real yang diperoleh tersebut disebut elemen, atau nilai, atau suku dari barisan tersebut. Untuk menulisakan elemen dari $R$ yang berpasangan dengan $n \in N$ biasanya dengan huruf kecil $x_n$, (atau $a_n$, atau $z_n$), sedangkan untuk menulisankan barisannya  kita gunakan huruf kapital X atau $X_n$, atau bisa menggunakan huruf kecil asalkan ditulis dalam kurung, yakni $(x_n: \ n \in N)$ atau $(x_n)$ saja. Penulisan  $(x_n)$ menyatakan bahwa urutan yang diwarisi dari $N$ adalah hal yang penting untuk membedakan penulisan dengan $\{x_n\}$. Contoh, $X=((-1)^n: \ n \in N)$ adalah barisan yang suku-sukunya mempunyai urutan yang berganti-ganti –1 dan 1, sedangkan himpunan nilai barisan tersebut adalah $\{(-1)^n: \ n \in N\}$ sama dengan {-1, 1}. Kita juga dapat menulisakan barisan dengan menulis suku-sukunya, dan berhenti setelah aturan formasinya kelihatan. Contoh, X=(2, 4, 6, 8, … ) yang merupakan barisan bilangan genap positif. Tapi, metode yang lebih memuaskan adalah menulis formula untuk suku umum dari barisan tersebut, yaitu $X=(2n: \ n \in N)$ atau $X=(2n)$. Selain daripada itu, ada juga pendefinisian barisan secara induktif atau rekursif, yaitu dengan menentukan nilai $x_1$  dan suatu formula terlebih dahulu untuk mendapatkan $x_{n+1} \ \  (n \ge 1)$ (bila $x_n$ diketahui) dan untuk mendapatkan formula $x_{n+1} \ \  (n \ge 1)$ dari $x_1$, $x_2$, … , $x_n$. Dengan cara ini, barisan bilangan genap di atas dapat kita definisikan dengan:

$x_1=2 \ \ \ \ x_{n+1}=x_n+2 \ \ (n \ge 1)$, atau

$x_1=2 \ \ \ \ x_{n+1}=x_1+x_n \ \ (n \ge 1)$.

Definisi Limit suatu Barisan: Misalkan $X=(x_n)$ barisan bilangan real. $X=(x_n)$ dikatakan konvergen ke $x \in R$, atau $x$ dikatakan limit dari $(x_n)$, bila untuk setiap $\epsilon >0$ terdapat bilangan asli $K(\epsilon)$, sedemikian sehingga untuk setiap $n \ge K(\epsilon)$, suku-suku $x_n$ memenuhi $|x_n – x| < \epsilon$.

Catatan: Bila suatu barisan tidak mempunyai limit, kita katakan divergen. Penulisan $K(\epsilon)$ digunakan untuk menunjukkan secara eksplisit bahwa pemilihan $K$ tergantung pada $\epsilon$, namun demikian sering lebih mudah menuliskan dengan $K$, daripada $K(\epsilon)$. Bila suatu barisan $X=(x_n)$ mempunyai limit di $x \in R$, kita gunakan notasi $\lim X=x$ atau $\lim (x_n)=x$ atau $\lim_{x \rightarrow \infty} x_ n =x$ atau $x_n \rightarrow x$.

Dengan definisi di atas, kita coba buktikan:

  • $\lim (\frac{1}{n})=0$


Ambil sebarang $\epsilon >0$, maka $\frac{1}{ \epsilon} >0$. Menurut sifat Archimedes, terdapat $K$ dimana $\frac{1}{K} < \epsilon$. Sehingga, untuk setiap $n \ge K(\epsilon)$ berlaku

$\begin{align} |\frac{1}{n} – 0| &= |\frac{1}{n}|\\ &=\frac{1}{n} \le \frac{1}{K} < \epsilon \end{align}$

Jadi, terukti $\frac{1}{n} \rightarrow 0$.

  • $\lim (3+ \frac{2}{n^2}) = 3$


Ambil sebarang $\epsilon >0$, maka $\frac{2}{\epsilon} >0$. Dengan sifat Archimedes kita dapat menunjukkan bahwa terdapat $K$ dimana $K >  \frac{2}{\epsilon}$. Sehingga, untuk setiap $n \ge K(\epsilon)$, berlaku:

$\begin{align} |3+ \frac{2}{n^2} – 3| &= | \frac{2}{n^2}| \\ &= \frac{2}{n^2} \\ & \le \frac{2}{K^2} < \frac{2}{K} < \frac{2}{\frac{2}{\epsilon}}=\epsilon \end{align}$

Jadi, terukti $3+ \frac{2}{n^2} \rightarrow 0$.

Yang sulit disini adalah memilih $K$ yang tepat. Sebaiknya, kita melakukan analisis pendahuluan terlebih dahulu sebelum masuk ke pembuktian formalnya. Misalnya, pada contoh 1:


$\begin{align} |\frac{1}{n} – 0| &= |\frac{1}{n}|\\ &=\frac{1}{n} \\ & \le \frac{1}{K}  < \epsilon \end{align}$

Kita harus memilih $K$ dimana $\frac{1}{K} < \epsilon$ (dijamin sifat Archimedes) agar $\frac{1}{K} < \epsilon$ .

Pada contoh 2:


$\begin{align} |3+ \frac{2}{n^2} – 3| &= |\frac{2}{n^2}| \\ &= \frac{2}{n^2} \\ & \le \frac{2}{K^2} \\ & < \frac{2}{K}  < \frac{2}{\frac{2}{\epsilon}}=\epsilon \end{align}$

Kita harus memilih $K$ dimana $K > \frac{2}{\epsilon}$ (dijamin dengan menggunakan sifat Archimedes)  agar $\frac{2}{K} < \frac{2}{\frac{2}{\epsilon}}=\epsilon $.

Catatan: Jika $a \ge b$ maka $\frac{1}{a} \le \frac{1}{b}$. Sehingga, jika $n \ge K$ maka:

  • $\frac{1}{n} \le \frac{1}{K}$
  • $\frac{2}{n^2} \le \frac{2}{K^2}$

Jadi, untuk membuktikan $x_n \rightarrow x$ menurut definisi, sebarang $\epsilon>0$ yang diberikan kita harus menemukan $K$ yang ada hubungannya dengan $\epsilon$  baik menggunakan hubungan “<” atau “>” secara “tepat” sehingga kita dapat menunjukkan bahwa untuk semua $n \ge K(\epsilon)$, suku-suku $x_n$ memenuhi $|x_n – x| < \epsilon$. Kita juga harus memastikan terdapatnya $K$ yang bergantung dengan $\epsilon$ tersebut.

Referensi: Buku Analisis Real karya Bartle dan Sherbert (edisi 4) dan terjemahannya.

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