Page 74 - E-MODUL FUNGSI DAN LIMIT DENGAN PENDEKATAN KONSTRUKTIVISME BERNUANSA PEMECAHAN MASALAH JOHN DEWEY

Nampak bahwa
1. lim ( ) = lim 2 = 2
→1 →1
2
2. lim ( ) = lim = 1
→1 →1
2
3. lim ( ) + ( ) = lim (2 + ) = 3
→1 →1
Sehingga berdasarkan fakta tersebut, ditulis

2
lim ( ) + ( ) = lim (2 + )
→1 →1
2
= lim 2 + lim
→1 →1
= lim ( ) + lim ( )
→1 →1
dapat dituliskan lim [ ( ) + ( )] = lim ( ) + lim ( )
→ → →

Teorema 5 [ ( ) − ( )] = ( ) − ( )
→ → →

Bukti:
lim [ ( ) − ( )] = lim [ ( ) + (−1) ( )]
→ →
= lim ( ) + lim (−1) ( )
→ →
= lim ( ) + (−1)lim ( )
→ →
= lim ( ) − lim ( )
→ →

Teorema 6 [ ( ) ∙ ( )] = ( ) ∙ ( )
→ → →
Andaikan nilai f(x) = 3x , g(x) = x + 2, serta sebaran nilai x disekitar 1 sehingga tabel

perhitungannya seperti berikut.

x 0,5 0,9 0,99 0,999 ... 1 ... 1,001 1,01 1,1 1,5
f(x) =3x 1,5 2,7 2,97 2,997 ... 3 ... 3,003 3,03 3,3 4,5

g(x) = x+2 2,5 2,9 2,99 2,999 ... 3 ... 3,001 3,01 3,1 3,5

f(x) ∙ g(x) 3,75 7,83 8,88 8,988 ... 9 ... 9,012 9,12 10,23 15,75

Terlihat bahwa
1. lim ( ) = lim 3 = 3
→1 →1
2. lim ( ) = lim + 2 = 3
→1 →1
3. lim [ ( ) ∙ ( )] = lim3 ( + 2) = 9
→1 →1

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