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Saturday, April 29, 2017

Definisi Limit Fungsi - Analisis Real

Definisi Limit

A. \lim_{x\rightarrow c}f(x)=L
Definisi :
\forall \epsilon >0,\exists \delta >0\ni |x-c|<\delta \rightarrow |f(x)-L|<\epsilon








B. \lim_{x\rightarrow c}f(x)=+\infty
Definisi :
\forall \beta >0,\exists \delta >0\ni |x-c|<\delta \rightarrow f(x)>\alpha








C. \lim_{x\rightarrow c}f(x)=-\infty
Definisi :
\forall \beta >0,\exists \delta >0\ni |x-c|<\delta \rightarrow f(x)<\beta  










D. \lim_{x\rightarrow c^+}f(x)=L
Definisi :
\forall \alpha >0,\exists \delta >0\ni x-c<\delta \rightarrow |f(x)-L|<\epsilon
  







E. \lim_{x\rightarrow c^+}f(x)=+\infty
Definisi :
\forall \alpha >0,\exists \delta >0\ni x-c<\delta \rightarrow f(x)> \alpha









F. \lim_{x\rightarrow c^+}f(x)=-\infty
Definisi :
\forall \beta >0,\exists \delta >0\ni x-c<\delta \rightarrow f(x)<\beta









G. \lim_{x\rightarrow \infty}f(x)=\infty
Definisi :
Misal : x\in (a,\infty) atau a<x<\infty, a>0
\forall \alpha \in R, \alpha >0,\exists k(\alpha)>a ,\exists jika jika x > k maka f(x) > α










H. \lim_{x\rightarrow \infty}f(x)=-\infty
Definisi :
Misal : x\in (a,\infty) atau a<x<\infty, a>0
\forall \beta \in R, \beta <0,\exists k(\beta)>a,\exists jika jika x > k maka f(x) < β









I. \lim_{x\rightarrow -\infty}f(x)=\infty
Definisi :
Misal : x\in (-\infty,b) atau -\infty<x<b, b<0
\forall \alpha \in R, \alpha >0,\exists k(\alpha)<b,\exists jika jika x < k maka f(x) > α










J. \lim_{x\rightarrow -\infty}f(x)=-\infty
Definisi :
Misal : x\in (-\infty,b) atau -\infty<x<b, b<0
\forall \beta \in R, \beta <0,\exists k(\beta)<0,\exists jika jika x < k maka f(x) > β











Contoh:
Buktikan bahwa \lim_{x\rightarrow 2} 3x+5=11

Jawab:
Analisis Pendahuluan
|x-c|\leq \delta maka |f(x)-L|\leq \epsilon
|x-c|\leq \delta \rightarrow |(3x+5)-11|\leq \epsilon
\rightarrow |3x-6|\leq \epsilon
\rightarrow 3|x-2|\leq \epsilon
\rightarrow |x-2|\leq \frac{\epsilon}{3}
\therefore \delta=\left \lceil \frac{\epsilon}{3}  \right \rceil

Bukti Formal
\forall \epsilon > 0, \exists \frac{\epsilon}{3}\ni jika |x-2|\leq \delta maka |(3x+5)-11|\leq \epsilon |x-2|\leq \frac{\epsilon}{3}
|(3x+5)-11|\leq \epsilon
|3x-6|\leq \epsilon
3|x-2|\leq \epsilon
3.\frac{\epsilon}{3} \rightarrow \epsilon \leq \epsilon
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