Page 37 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 35

[ដំេ�ះ��យ]

I. គណនាលីមីត៖
2
3
x – x + x – 1
0
ក. lim (មានរាងមិនកំណត់ )
x→1 x – 1 0
2
2
(x – 1)x + (x – 1) (x – 1)(x + 1)
2
2
= lim = lim = lim(x + 1) = 1 + 1 = 2
x→1 (x – 1) x→1 (x – 1) x→1
3
2
x – x + x – 1
lim = 2
x→1 x – 1
2
sin x – 1
0
ខ. lim (មានរាងមិនកំណត់ )
x→– 1 + sin x 0
π
2
) (
( sin x – 1 sin x + 1 ) ( )
(
)
= lim = lim sin x – 1 = sin – π – 1 = –1 – 1 = –2
x→– π sin x + 1 x→– π 2
2 2
2
sin x – 1
lim = –2
x→– 1 + sin x
π
2
( √ )
2
គ. lim x + x – x (មានរាងមិនកំណត់+∞ – ∞)
x→+∞
( ) ( )
√ √
2
2
x + x – x x + x + x x + x – x 2
2
= lim ( ) = lim √
x→+∞ √ 2 x→+∞ 1
x + x + x x 1 + + x
x
x 1 1 1
= lim ( √ ) = lim √ = =
x→+∞ 1 x→+∞ 1 1 + 1 2
x 1 + + 1 1 + + 1
x x
)
( √ 1
2
lim x + x – x =
x→+∞ 2
( –x x ) 2
e + e sin x
0
ឃ. lim (មានរាងមិនកំណត់ )
x→0 2x 2 0
0
2
2
e –x + e x sin x e + e 0 2 ( e –x + e x ) sin x
= lim · = × 1 = = 1 lim = 1
x→0 2 x 2 2 2 x→0 2x 2
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