Page 17 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 15

[ដំេ�ះ��យ]

I. គណនាលីមីត៖
1 – x 2
0
ក. lim (មានរាងមិនកំណត់ )
2
x→1 x + 2 – 3x 0
(1 – x)(1 + x) –(x – 1)(1 + x) –(1 + x) –(1 + 1)
= lim = lim = lim = = 2
x→1 (x – 1)(x – 2) x→1 (x – 1)(x – 2) x→1 x – 2 1 – 2

1 – x 2
ដូច ះ lim = 2
2
x→1 x + 2 – 3x
√
x + 6 – 3
0
ខ. lim (មានរាងមិនកំណត់ )
0
3
x→3 x – 27
( √ ) ( √ )
x + 6 – 3 x + 6 + 3 x + 6 – 9
= lim ( ) ( √ ) = lim ( ) ( √ )
3
x→3 x – 3 3 x + 6 + 3 x→3 (x – 3) x + 3x + 9 x + 6 + 3
2
1 1 1
= lim ( ) ( √ ) = ( √ ) =
2
x→3 x + 3x + 9 x + 6 + 3 (9 + 9 + 9) 9 + 3 162
√
x + 6 – 3 1
ដូច ះ lim =
3
x→3 x – 27 162
5 sin 5x
0
គ. lim (មានរាងមិនកំណត់ )
0
x→0 x
sin 5x 5 sin 5x
= lim 25 · = 25 × 1 = 25 ដូច ះ lim = 25
x→0 5x x→0 x
II. គណនា z 1 + z 2 , (z 1 + z 2 ) × z 3
√ ( √ ) ( √ ) 1
ើងមាន z 1 = 3 – i ; z 2 = 1 – 3 + 1 – 3 i ; z 3 = –
2
√ √ √ √ √
z 1 + z 2 = 3 – i + 1 – 3 + i – 3i = 1 – 3i ដូច ះ z 1 + z 2 = 1 – 3i
√ √
( )
( √ ) 1 1 3 1 3
( ) ( )
z 1 + z 2 × z 3 = 1 – 3i – = – + i ដូច ះ z 1 + z 2 × z 3 = – + i
2 2 2 2 2

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