MATEMATIKA 1 skripta studij: Biotehnologija i Prehrambena ... - PBF

Zadatak 159 Izračunajte lim x→1 lnx·ln(x−1).
Rješenje:
ln(x−1)
limlnx·ln(x−1) = |0·(−∞)| = lim
x→1 x→1
1
lnx
∣
1
x−1
x→1 − 1 · 1
ln 2 x x
= lim
= −lim
x→1
ln 2 x
∣ ∣∣∣
x−1 = 0
0∣ = −lim
x→1
= ∣ ∞ ∣
∞
2lnx· 1
x
1
= 0.
Zadatak 160 Izračunajte lim x→π/2
(
Rješenje:
lim
x→π/2
x
− π
ctgx 2cosx
)
.
( x
ctgx − π )
= |∞−∞| = lim
2cosx
x→π/2
2xsinxcosx−πcosx
= lim
x→π/2 2cos 2 x
= 1 2 lim 2sinx+2xcosx
x→π/2 −sinx
2xcosx−π cosx
sinx
cosx
·2cosx sinx =
0
∣0∣
= 1 2 lim 2xsinx−π
x→π/2 cosx
= 1 2 · 2·1+2· π
2 ·0
−1
= −1.
Zadatak 161 Izračunajte lim x→1
( x
x−1 − 1
lnx)
.
Rješenje:
lim
x→1
( x
x−1 − 1
lnx
= lim
x→1
)
∣ xlnx−x+1 ∣∣∣
= |∞−∞| = lim
x→1 (x−1)lnx = 0
0∣ = lim
x→1
∣ lnx ∣∣∣
xlnx+(x−1) = 0
0∣ = lim
x→1
1
x
1+lnx+1 = 1
1+1 = 1 2 .
lnx
lnx+ x−1
x
Zadatak 162 Izračunajte a) lim x→∞
lnx
x
b) lim x→∞ (x−lnx).
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