Page 76 - BUKU MATEMATIKA DASAR - KALKULUS DIFERENSIAL

5. lim f (x )  g (x ))  lim f (x )  lim g (x )
(
x c x c x c
(
6. lim f (x ) g (x ))  lim f (x ) lim g (x )
x c x c x a
f (x ) lim f (x ) 0
7. lim  x c , asalkan lim g (x ) 
x c g (x ) lim g (x ) x c
x c
n
n
8. lim ( f ( x))  lim f ( x) 
x c x c
Teorema di atas, dapat diaplikasikan dalam banyak hal pada

penyelesaian soal-soal tentang limit.

Contoh :

1. Jika lim f (x )  3 dan lim g (x )   1
x a x a
Tentukan :

a. lim f 2 (x )  g 2 (x )  ...
x a


lim f 2 (x ) g 2 (x )  lim { f 2 (x ) g 2 (x )
x a x a

 lim f 2 (x )  lim g 2 (x )
x a x a

 lim f (x )  lim g (x  )
2
2
x a x a
 3  ( ) 1
2
2
 10

 
b. lim 3 f ( x)  xg )(  3  lim 3 f ( x)  lim g( x)   3
x a x a  x a 
 lim 3 f (x ) { lim g (x )  lim } 3
x a x a x  3
 3 lim f (x ) { lim g (x )  lim } 3
x a x a x a
 3 ( 3  1 ) 3

 2 3 3

2 f (x )  3g (x ) lim 2 f (x )  3g (x  )
c. lim  x 2
x a f (x )  g (x ) lim  (xf )  g (x  )
x 2
lim 2 f (x )  lim 3g (x )
 x 2 x 2
lim f (x )  lim g (x )
x a x a
2 lim f (x )  3 lim g (x
)
 x 2 x 2
lim f (x )  lim g (x )
x a x a
) 3 ( 2  ( 3  ) 1

3  ( ) 1
9

2