Arctan .se
Exempel 1
Ber?akna f?oljande gr?ansv?arden:
a lim x0
1 arctan
x
-
1 ln(1 +
x)
b
lim
x
xe1/x
-
x2
sin
1 x
c
cos x - |1 - x2| + x7
lim
x0
x4 + x7
d
lim
x
cos
x
-
x4
|1 - x2| + x7
+
x7
1/8
lim
x0
1 arctan
x
-
ln
1 (1 +
x)
=?
Maclaurinutveckling t.o.m. ordning 2, d v s med restterm av grad 3 ger
ln
(1
+
x)
=
x
-
1 2
x2
+
O
x3
arctan x = x + O x3
=
ln (1
+
x) - x2
arctan x
=
x
-
1 2
x2
+
O(x3) - x2
(x
+
O(x3))
=
-
1 2
x2
+
O(x3)
x2
=
-
1 2
+ O(x)
-
1 2
d?a x 0.
3/8
lim
x0
1 arctan
x
-
ln
1 (1 +
x)
=?
Vi b?orjar med att s?atta p?a gemensamt br?ak och utnyttja standardgr?ansv?arden f?or att slippa utveckla s?a m?anga termer.
lim
x0
1 arctan
x
-
ln
1 (1 +
x)
=
lim
x0
ln (1 + x) arctan x
- ln
arctan x (1 + x)
=
=
lim
x0
ln (1 + x) - arctan x
arctan x
x
?
ln
(1 + x
x)
?
x2
=
r?akneregler f?or gr?ansv?arden
=
=
lim
x0
ln
(1
+
x) - x2
arctan
x.
G?or vi p?a detta s?att f?ar vi ocks?a direkt svar p?a hur l?angt vi beh?over utveckla termerna i t?aljaren, i detta fall till grad 2.
2/8
lim
x
xe1/x
-
x2
sin
1 x
=?
S?att
t
=
1 x
.
Observera
att
d?a x
s?a t
0+
eftersom
1 x
>
0.
Vi kan ju bara g?a mot fr?an ett h?all. Vi f?ar
xe1/x
=
1 t
et
=
1 t
1
+
t
+
1 2
t2
+
1 6
t3
+
O
t4
=
=
1 t
+ 1 + O(t) ,
x2 sin
1 x
=
1 t2
sin t
=
1 t2
t
-
1 6
t3
+
O
t5
=
=
1 t
+ O(t) .
D?a det r?acker med en nollskild term och restterm ser vi att kan stanna utvecklingarna redan vid O(t).
4/8
lim
x
xe1/x
-
x2
sin
1 x
=?
Vi f?ar
lim
x
xe1/x
-
x2
sin
1 x
=
t = 1/x x t 0+
=
= lim t0+
1 t
et
-
1 t2
sin
t
= lim t0+
1 t
+
1
+
O(t)
-
1 t
+
O(t)
=
= lim (1 + O(t)) = 1 t0+
5/8
cos x - |1 - x2| + x7
lim
x0
x4 + x7
Ins?attning av utvecklingarna ger t?aljaren
T : cos x- |1-x2|+x7 =
=
1-
1 2
x2+
1 24
x4
+O
x6
-
1-
1 2
x2
-
1 8
x4+O
x6
=
1 24
+
1 8
x4+O x6
+x7
=
1 6
x4+O
x6
N : x4 + x7
T N
=
1 6
x4+O(x6)
x4 + x7
=
1 6
+O(x2)
1 + x3
1 6
d?a x 0
+x7 =
7/8
lim
x0
cos
x
-
x4
|1 - x2| + x7
+
x7
D?a x 0 kan vi ta bort beloppet innanf?or rottecknet; om x 0 s?a m?aste ju x2 f?orr eller senare vara mindre ?an 1. Vidare, d?a x ?ar litet dominerar x4 ?over x7 i n?amnaren. F?or att vi d?a skall kunna
f?a ut n?at ur detta m?aste vi utveckla t.o.m. ordning 4. Vi f?ar
cos x = 1 - 1 x2 + 1 x4 + O x6 , 2 24
1 - x2 =
1 - x2 1/2 =
t?ank t = -x2
=
=
1
+
1 2
(-x2)
+
1/2 2
-x2 2 + O -x2 3 =
=
1/2
=
1 2
(
1 2
- 1)
= -1
= 1 - 1 x2 - 1 x4 + O x6
2
2
8
28
6/8
cos x - |1 - x2| + x7
lim
x
x4 + x7
Kuggfr?aga! H?or ju hemma i Envariabelanalys, del 1. Bryt ut x2 ur rotuttrycket och x7 i t?aljare och n?amnare eftersom det ?ar den som
?ar dominant d?a x ?ar stort. Vi f?ar
cos x - |1 - x2| + x7 x4 + x7
=
cos x x7
+
1-
1 x2
x6
1
+
1 x3
+1
1
d?a x .
8/8
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