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Cara Membuktikan Nilai Limit Menggunakan Definisi



Sebelum kita membuktikan nilai limit suatu fungsi dengan menggunakan definisi, tentu kita harus mengetahui definisnya terlebih dahulu. Limit $f(x)$ untuk $x$ mendekati $c$ adalah $L$ ditulis $\lim_{x \rightarrow c} f(x)=L$ jika dan hanya jika:
Untuk setiap $\epsilon >0$ terdapat $\delta >0$ sedemikian sehingga jika $0 <|x-c|< \delta$ maka $|f(x)-L|<0$
Sebelum kita menggunakan definisi di atas, kita harus benar-benar memahami arti dari setiap kata dalam definisi tersebut agar kita dapat memahami definisi tersebut dengan benar. Limit dalam bahasa sehari-hari artinya batas. L adalah limit suatu fungsi f(x) maksudnya adalah batas dari nilai-nilai yang dihasilkan apabila kita memberikan nilai-nilai x yang mendekati nilai c ($x \neq c$) baik dari arah kiri maupun dari arah kanan. 

Misalnya, fungsinya adalah $f(x)=x-2$ maka nilai limitnya adalah 1 untuk nilai-nilai x yang mendekati 3; nilai limitnya adalah 2 untuk nilai-nilai x yang mendekati 4; nilai limitnya adalah -1 untuk nilai-nilai x yang mendekati 1. Apabila nilai-nilai yang diberikan semakin dekat dengan c yang dimaksud maka semakin jelas bahwa f(x) mendekati nilai L. Untuk itu jika L betul-betul limit dari f(x) untuk x mendekati c maka seharusnya, seberapa kecil pun jarak yang diberikan antara f(x) dengan L yang disimbolkan dengan $\epsilon$ maka pasti derdapat $\delta$ yang berpadanan dimana $\delta$ ini adalah jarak antara x dan c sehingga apabila $0<|x-c|< \delta$ maka mengakibatkan $|f(x)-L|< \epsilon$. Ingat bahwa tanda mutlak itu menyatakan jarak dan karena x tidak mempersyaratkan harus sama dengan c maka ditulis $0<|x-c|< \delta$. 

Oleh karena itu, untuk membuktikan nilai limit suatu fungsi $f(x)$ adalah L untuk $x$ mendekati suatu titik $c$ dengan menggunakan definisi maka kita harus menunjukan keberadaan $\delta >0$ jika diberikan sebarang $\epsilon >0$ dan menunjukkan bahwa apabila $0<|x-c|< \delta$ maka mengakibatkan $|f(x)-L| < \epsilon$.

Contoh:

Buktikan bahwa $\lim_{x \rightarrow 4} 3x-7 =5$

Analisis pendahuluan: Andaikan $\epsilon$ bilangan positif sebarang. Kita harus menghasilkan $\delta >0$ sedemikian sehingga jika $0 <|x-4|< \delta$ maka $|(3x-7)-5| < \epsilon$. Pandang ketaksamaan di sebelah kiri:

$\begin{align} |(3x-7)-5| < \epsilon ⇔ |3x-12| & < \epsilon \\ |3(x-4)| & < \epsilon \\ |3||x-4|  & < \epsilon \\ |x-4| & < \frac{ \epsilon}{3} \end{align}$

Sekarang kita lihat bagaimana memilih $\delta$, yakni $\delta=  \frac{ \epsilon}{3}$ atau $\delta <  \frac{ \epsilon}{3}$ . Tetu saja sebarang $\delta$ yang lebih kecil akan memenuhi. Sekarang kita tulis buktinya secara formal berikut ini.

Andaikan diberikan $\epsilon >0$ pilih $\delta =  \frac{ \epsilon}{3}$ sedemikian hingga jika $0 <|x-4|< \delta$ maka
$\begin{align} |(3x-7)-5| &= |3x-12| \\ &=|3(x-4)| \\ &= 3|x-4| \\ & < 3 \delta = \epsilon \end{align}$

Atau bisa juga seperti di bawah ini.

Andaikan diberikan $\epsilon >0$ pilih $\delta <  \frac{ \epsilon}{3}$ sedemikian hingga jika $0 <|x-4|< \delta$ maka
$\begin{align} |(3x-7)-5| &= |3x-12| \\ &=|3(x-4)| \\ &= 3|x-4| \\ & < 3 \delta < \epsilon \end{align}$

Jadi, terbukti bahwa $\lim_{x \rightarrow 4} 3x-7 =5$.

Referensi: Kalkulus 1 Purcell

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