Page 48 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 46

[ដំេ�ះ��យ]

I. គណនាលីមីតខាង ម៖
( )
2 sin x – π 4
0
ក. lim (មានរាងមិនកំណត់ )
π π – x 0
x→ 4 4
( ) ( )
sin x – π 2 2 sin x – π
= lim 2 ( 4 ) = × 1 = –2 lim π 4 = –2
x→ π – x – π –1 x→ π – x
4 4 4 4
–2 sin 5x
0
ខ. lim √ √ (មានរាងមិនកំណត់ )
0
x→0 5 – x + 5
( √ √ ) ( √ √ )
–2 sin 5x 5 + x + 5 –2 sin 5x 5 + x + 5
= lim ( √ √ ) ( √ √ ) = lim
x→0 5 – x + 5 5 + x + 5 x→0 5 – (x + 5)
sin 5x √ √ ) √ √
(
= lim 2 × 5 5 + x + 5 = 10 × 1 × 2 5 = 20 5
x→0 5x
–2 sin 5x √
lim √ √ = 20 5
x→0 5 – x + 5
2
1 – cos 3x
0
គ. lim (មានរាងមិនកំណត់ )
x→0 –2x 2 0
2
2
(
sin 3x 3 2 ) ( 9 ) 9 1 – cos 3x 9
= lim = 1 × – = – lim = –
x→0 (3x) 2 –2 2 2 x→0 –2x 2 2
2
x – x
0
ឃ. lim (មានរាងមិនកំណត់ )
x→0 |x| 0
 x(x – 1) 0 – 1
 lim = = 1

2

x(x – 1)  x→0 – –x –1 x – x


= lim =    lim = ±1
x→0 |x|    x(x – 1) 0 – 1 x→0 |x|
 lim = = –1

x→0 + x 1
II. ក. គណនា z 1 + z 2 , z 1 – z 2 , z 1 z 2
√ √
យ z 1 = –1 + i 3, z 2 = –1 – i 3 បាន
√ ( √ )
• z 1 + z 2 = –1 + i 3 + –1 – i 3 = –2 z 1 + z 2 = –2
√ ( √ ) √ √
• z 1 – z 2 = –1 + i 3 – –1 – i 3 = 2 3i z 1 – z 2 = 2 3i
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