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DESCRIPTION
fffTRANSCRIPT
RJESENJA 2. SKOLSKE ZADACE IZ MATEMATIKE1
13.11.2006. grupe 1,3,5
A grupa
1. limn→∞
3n + 7
8n− 2
5n
= limn→∞
3
8
5n
·n + 7
3
n− 1
4
5n
= limn→∞
3
8
5n
·
1+ 112n−3
31
12n−3
31 ·
31
12n−3·5n
= 0 · e31·5
12 = 0.
2. limn→∞
(√
x4 + 2x2−√
x4 + 1)·
√ x4 + 2x2 +
√ x4 + 1
√ x4 + 2x2 +
√ x4 + 1
= limn→∞
x4 + 2x2 − x4 − 1√
x4 + 2x2 +√
x4 + 1
= limn→∞
2x2 − 1√
x4 + 2x2 +√
x4 + 1
/ : x2
/ : x2 = lim
n→∞
2− 1
x2
1 + 2
x2 +
1 + 1
x4
= 2
1 + 1 = 1.
3. limn→−∞
arcsin e1
x = arcsin( limn→−∞
e1
x ) = arcsin e0 = arcsin 1 = π
2.
4. 2 = f (0) = limx→0
f (x) = limx→0
sin ax
3x = lim
x→0
sin ax
ax ·
a
3 = 1 ·
a
3 =
a
3 ⇒ a = 6.
B grupa
1. limn→∞
2n + 3 · 5n
3n − 4n= lim
n→∞
2n + 3 · 5n
4n
3
4
n
− 1
/ : 5n
/ : 5n= lim
n→∞
2n + 3 · 5n
3n − 4n= lim
n→∞
2
5
n
+ 3
4
5
n
·
3
4
n
− 1
= 0 + 3
0 · (0− 1) = −∞.
2. limx→∞
x + 1
x− 1
x−1
= limx→∞
1+ 1x−1
2
x−1
2 ·2
= e2.
3. limx→3+
th 3
x2 − 9 = th
limx→3+
3
x2 − 9
= th(+∞) = 1,
limx→3−
th 3
x2 − 9 = th
limx→3−
3
x2 − 9
= th(−∞) = −1 ⇒ LIMES NE P OSTOJI !
4. a = f (0) = limx→0
e− 1
x2 = lim
x→0
1
e 1
x2
= 1
e∞= 0.