Page 95 - BacII 2011-2017 by Lim Seyha

វិទយល័យសេម�ចឳ េខត�េស ម�ប 93

[ដំេ�ះ��យ]

I. គណនាលីមីត៖
3
3
3
( ) ( )
ក. lim 3x – 4x = 3(1) – 4(1) = 3 – 4 = –1 ដូច ះ lim 3x – 4x = –1
x→1 x→1
3
x – 8 0
ខ. lim √ ( មានរាងមិនកំណត់ 0 )
x→2 x + 2 – 2
( ) ( √ ) ( ) ( √ )
3
2
x – 2 3 x + 2 + 2 (x – 2) x + 2x + 4 x + 2 + 2
= lim ( √ ) × ( √ ) = lim
x→2 x + 2 – 2 x + 2 + 2 x→2 (x + 2) – 4
( ) ( √ ) ( ) ( √ )
2
2
= lim x + 2x + 4 x + 2 + 2 = 2 + 2 · 2 + 4 2 + 2 + 2 = 48
x→2
3
x – 8
ដូ ះ lim √ = 48
x→2 x + 2 – 2
( )
គ. lim ln x – x 2 (មានរាងមិនកំណត់ +∞ – ∞)
x→+∞
( ln x )
= lim x 2 – 1 = +∞(0 – 1) = –∞
x→+∞ x 2
( 2 )
ដូច ះ lim ln x – x = –∞
x→+∞
II. គណនាអាំង ល
∫ [ 3 ] 2
2 ( ) x ( )
3
2
ក. I = 1 – 3x dx = x – 3 = 2 – 2 – 1 – 1 3 = 2 – 8 = –6 ដូច ះ I = –6
1 3 1
3 1 3 x –2+1 1 1 1
∫ ∫ [ ] 3 [ ] 3 ( )
–2
J = dx = x dx = = – = – – –
2 x 2 2 –2 + 1 2 x 2 3 2
ខ.
–2 + 3 1 1
= = ដូច ះ J =
6 6 6
1 1
∫ ( )
គ. K = – 1 dx = ln |x + e| – x ] 1 0 = ln |1 + e| – 1 – ln |0 + e| – 0 )
(
[
0 x + e
= ln(1 + e) – 1 – ln e = ln(1 + e) – 1 – 1 = ln(1 + e) – 2
ដូច ះ K = ln(1 + e) – 2
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