ANREL_TEOREMA LIMIT.pptx

TEOREMA LIMIT
1. Miftakhul Khoiroh (2420002)
2. Nur Afifah Muzaidah (2420005)
3. Nanda Nisaul Maghfiroh
4. Fitria Alfiatus Sa’adah
5. Ahmad Habibulloh
Oleh kelompok 3:

4.2.1 Definisi
Misalkan 𝐴 ⊆ ℝ , misalkan 𝑓 ∶ 𝐴 → ℝ dan
misalkan 𝑐 ∈ ℝ adalah titik cluster dari A. Kita
katakan bahwa 𝑓 dibatasi pada lingkungan 𝑐 jika
terdapat 𝛿-keliling 𝑉𝛿(𝑐) dari c dan a konstanta
𝑀 > 0 sehingga kita memiliki 𝑓 𝑥 ≤ 𝑀 untuk
semua 𝑥 ∈ 𝐴 ∩ 𝑉𝛿 𝑐 .

Bukti. Jika 𝐿 ∶= lim
𝑥→𝑐
𝑓 ,maka untuk 𝜀 = 1, terdapat 𝛿 > 0 sehingga jika
0 < 𝑥 − 𝑐 < 𝛿 ,sehingga 𝑓 𝑥 − 𝐿 < 1 ;
( oleh Corollary 2.2.4(a))
𝑓 𝑥 − 𝐿 ≤ 𝑓 𝑥 − 𝐿 < 1
Oleh karena itu, jika 𝑥 ∈ 𝐴 ∩ 𝑉𝛿 𝑐 , 𝑥 ≠ 𝑐 ,maka 𝑓 𝑥 ≤ 𝐿 + 1 .Jika
𝑐 ∉ 𝐴, kita ambil 𝑀 = 𝐿 + 1 ,sedangkan jika 𝑐 ∈ 𝐴 kita ambil
𝑀 ∶= sup 𝑓 𝑐 , 𝐿 + 1 . Olehh karena itu jika 𝑥 ∈ 𝐴 ∩ 𝑉𝛿 𝑐 ,
maka 𝑓 𝑥 ≤ 𝑀 .Hal ini menunjukkan bahwa 𝑓 dibatasi pada
lingkungan 𝑉𝛿 𝑐 dari 𝑐.
.
4.2.2 Teorema Jika 𝐴 ⊆ ℝ dan 𝑓 ∶ 𝐴 → ℝ memiliki limit di
𝑐 ∈ ℝ , maka 𝑓 dibatasi pada suatu lingkungan dari c.

Misalkan 𝐴 ⊆ ℝ dan misalkan 𝑓 dan 𝑔 adalah fungsi yang didefinisikan
pada 𝐴 ke ℝ. Kita definisikan jumlah 𝑓 + 𝑔, selisih 𝑓 − 𝑔 dan hasil kali
𝑓𝑔 pada 𝐴 ke ℝ menjadi fungsi yang diberikan oleh
𝑓 + 𝑔 𝑥 ∶= 𝑓 𝑥 + 𝑔 𝑥 , 𝑓 − 𝑔 𝑥 ∶= 𝑓 𝑥 − 𝑔 𝑥 ,
𝑓𝑔 𝑥 ∶= 𝑓 𝑥 𝑔 𝑥
untuk semua 𝑥 ∈ 𝐴. Selanjutnya, jika 𝑏 ∈ ℝ, kita mendefinisikan
kelipatan 𝑏𝑓 sebagai fungsi yang diberikan oleh
𝑏𝑓 𝑥 ∶= 𝑏𝑓 𝑥 untuk semua 𝑥 ∈ 𝐴.
Akhirnya, jika ℎ 𝑥 ≠ 0 untuk 𝑥 ∈ 𝐴, kita mendefinisikan hasil bagi 𝑓/ℎ
sebagai fungsi yang diberikan oleh
𝑓
ℎ
𝑥 ∶=
𝑓 𝑥
ℎ 𝑥
untuk semua 𝑥 ∈ 𝐴
4.2.3 Definisi

.
4.2.4 Teorema
Misalkan 𝐴 ⊆ ℝ, misalkan 𝑓 dan 𝑔 fungsi pada 𝐴 ke ℝ, dan
misalkan 𝑐 ∈ ℝ adalah titik cluster dari 𝐴. Selanjutnya, misalkan 𝑏 ∈ ℝ
(a) Jika lim
𝑥→𝑐
𝑓 = 𝐿 dan lim
𝑥→𝑐
𝑔 = 𝑀, maka:
lim
𝑥→𝑐
𝑓 + 𝑔 = 𝐿 + 𝑀, lim
𝑥→𝑐
𝑓 − 𝑔 = 𝐿 − 𝑀
lim
𝑥→𝑐
𝑓𝑔 = 𝐿𝑀, lim
𝑥→𝑐
𝑏𝑓 = 𝑏𝐿.
(b) Jika ℎ ∶ 𝐴 → ℝ, jika ℎ(𝑥) ≠ 0 untuk semua 𝑥 ∈ 𝐴, dan jika
lim
𝑥→𝑐
ℎ = 𝐻 ≠ 0, maka
lim
𝑥→𝑐
𝑓
ℎ
=
𝐿
𝐻
.

Bukti Salah satu pembuktian teorema ini sama persis dengan Teorema 3.2.3 Atau dapat dibuktikan dengan
menggunakan Teorema 3.2.3 dan 4.1.8. Sebagai contoh, misalkan (xn) adalah sembarang barisan di A
sedemikian sehingga xn ≠ c untuk 𝑛 ∈ 𝑁 dan c = lim(xn). Ini mengikuti dari Teorema 4.1.8 bahwa
lim(f(xn)) = L , lim(g(xn)) = M
Di samping itu. Definisi 4.2.3 menyiratkan bahwa
(fg)(xn) = f(xn)g(xn) untuk 𝑛 ∈ 𝑁
Oleh karena itu penerapan Teorema 3.2.3 menghasilkan
Lim ((fg)(xn)) = lim(f(xn)g(xn))
=[lim(f(xn))] [lim(g(xn))] = LM.
Akibatnya, berikut dari Teorema 4.1.8 bahwa
lim
𝑥→𝑐
= lim(fg)(xn)) = LM
Keterangan Biarkan 𝐴 ∁ 𝑅. dan biarkan f1,f2,...,fn. menjadi fungsi pada A ke R, dan misalkan c adalah titik
cluster dari A. Jika Lk
∶= lim
𝑥→𝑐
𝑓k untuk k = 1,...,n, kemudian mengikuti dari Teorema 4.2.4 dengan Argumen
induksi bahwa
L1 + L2 +...+ Ln = lim
𝑥→𝑐
(f1 +f2 +...+fn),
dan
L1 . L2 …
Ln = lim (f1 . f2 …
fn).
Secara khusus, kami menyimpulkan bahwa jika L = lim
𝑥→𝑐
𝑓 dan n ∈ N, maka
𝐿𝑛
= lim
𝑥→𝑐
𝑓 𝑥
𝑛

4.2.5 Contoh
(a) Beberapa limit yang ditentukan dalam Bagian 4.1 dapat dibuktikan dengan
menggunakan Teorema 4.2.4 Sebagai contoh, dari hasil ini diperoleh bahwa
sejaklim
𝑥→𝑐
𝑥 = 𝑐. lalu lim
𝑥→𝑐
𝑥
2
= 𝑐
2. dan jika 𝑐 > 0, maka
lim
𝑥→𝑐
1
𝑥
=
1
lim
𝑥→𝑐
x =
1
𝑐
.
(b) lim
𝑥→2
(𝑥2
+ 1)(𝑥3
− 4) = 20.
Ini mengikuti dari Teorema 4.2.4 bahwa
lim
𝑥→2
𝑥2
+ 1 𝑥3
− 4 = lim
𝑥→2
𝑥
2
+ 1 lim
𝑥→2
𝑥
3
− 4
= 5 . 4 = 20.
.
(c) lim
𝑥→2
𝑥3−4
𝑥2−1
=
4
5

Bukti. Jika 𝐿 = lim
𝑥→c
𝑓, maka menurut Teorema 4.1.8 bahwa
jika (𝑥𝑛) sebarang barisan bilangan real sedemikain
sehingga 𝑐 ≠ 𝑥𝑛 ∈ 𝐴 untuk semua 𝑛 ∈ ℕ dan jika barisan
(𝑥𝑛) konvergen ke c, maka barisan (𝑓(𝑥𝑛)) konvergen ke L.
Karena 𝑎 ≤ 𝑓(𝑥𝑛) ≤ 𝑏 untuk semua 𝑛 ∈ ℕ, berarti
menurut Teorema 3.2.6 bahwa 𝑎 ≤ 𝐿 ≤ 𝑏.
4.2.6 Teorema Misalkan 𝐴 ⊆ ℝ, 𝑓 ∶ 𝐴 → ℝ dan 𝑐 ∈ ℝ suatu titik
cluster dari A. Jika 𝑎 ≤ 𝑓(𝑥) ≤ 𝑏 untuk semua 𝑥 ∈ 𝐴, 𝑥 ≠ 𝑐, dan
jika lim
𝑥→c
𝑓 ada, maka 𝑎 ≤ lim
𝑥→c
𝑓 → ≤ 𝑏.

4.2.7 Teorema Apit
Misalkan 𝐴 ⊆ ℝ, misal 𝑓, 𝑔, ℎ ∶ 𝐴 → ℝ, dan 𝑐 ∈ ℝ suatu titik
cluster dari A. Jika 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) untuk semua 𝑥 ∈
𝐴, 𝑥 ≠ 𝑐, dan jika lim
𝑥→c
𝑓 = 𝐿 = lim
𝑥→c
ℎ, maka lim
𝑥→c
𝑔 = 𝐿.

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