ZnajdΕΊ wszystkie asymptoty funkcji: ...
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π
ZnajdΕΊ wszystkie asymptoty funkcji: π(π₯) = 2π₯ + 1 β π₯ (ππππ‘π (π₯) + 2 ) 1. OkreΕlamy dziedzinΔ funkcji π: π₯ β β \ {0} 2. Badamy czy dla π₯ = 0 funkcja ma asymptoty pionowe jednostronne: 1 1 π 1 1 π lim+ (2π₯ + 1 β (ππππ‘π ( ) + ) ) = lim+2π₯ + lim+1 β lim+ ( ) β lim+ (ππππ‘π ( ) + ) π₯β0 π₯β0 π₯β0 π₯β0 π₯ π₯ 2 π₯ π₯β0 π₯ 2 1 1 π π = 2 β 0+ + 1 β + β (ππππ‘π ( + ) + ) = 0 + 1 β β β (ππππ‘π(β) + ) = 1 β β β π = ββ 0 0 2 2 βΉ πππ π₯ = 0 ππ’πππππ ππ ππ π¦πππ‘ππ‘Δ ππππππ€Δ
ππππ€ππ π‘ππππΔ
1 1 π 1 1 π limβ (2π₯ + 1 β (ππππ‘π ( ) + ) ) == limβ2π₯ + limβ1 β limβ ( ) β limβ (ππππ‘π ( ) + ) π₯β0 π₯β0 π₯β0 π₯β0 π₯ π₯ 2 π₯ π₯β0 π₯ 2 1 1 π π π π = 0 + 1 β β β (ππππ‘π ( β ) + ) = 1 β (ββ) β (ππππ‘π(ββ) + ) = 1 + β β (β + ) 0 0 2 2 2 2 1 π ππππ‘π (π₯ ) + 2 1 1 π = 1 + [β β 0] = 1 β limβ ( ) β limβ (ππππ‘π ( ) + ) = 1 β limβ 1 π₯β0 π₯β0 π₯ π₯β0 π₯ 2 1 π₯ 1 π ππππ‘π (π₯ ) + 2 1 1 π 1 β limβ ( ) β limβ (ππππ‘π ( ) + ) = 1 β limβ 1 π₯β0 π₯β0 π₯ π₯β0 π₯ 2 1 π₯ 1 1 2 β (β π₯ 2 ) 1 π 1 ππππ‘π (π₯ ) + 2 π» 1 + (π₯ ) β1 = 1 β limβ β 1 β limβ = 1 β limβ 2 π₯β0 π₯β0 π₯β0 π₯ + 1 π₯ 1 1 1 β (β 2 ) π₯ 1 π 1 2 ππππ‘π ( ) + π» 1+( ) β1 π₯ 2 π₯ = 1 β limβ β 1 β limβ = 1 β limβ 2 π₯β0 π₯β0 π₯β0 π₯ + 1 π₯ 1 β1 =1β β 2 = 1 β (β1) = 2 βΉ ππππ ππ π¦πππ‘ππ‘π¦ ππππππ€ππ πππ€ππ π‘ππππππ πππ π₯ (0 ) + 1 β1 =1β β 2 = 1 β (β1) = 2 βΉ ππππ ππ π¦πππ‘ππ‘π¦ ππππππ€ππ πππ€ππ π‘ππππππ πππ π₯ (0 ) + 1 =0 =0 3. Szukamy asymptot poziomych lub ukoΕnych w nieskoΕczonoΕci: a.
π(π₯) π₯β+β π₯
lim
= lim
1 π₯
1 π₯
π 2
2π₯+1β (ππππ‘π( )+ ) π₯
π₯β+β
2π₯ π₯β+β π₯
= lim 1
1 π₯β+β π₯
+ lim 1
β lim
π₯β+β
1 π₯ π₯2
ππππ‘π( )+
π 2
π
=2 1
1
lim [π(π₯) β 2π₯)] = lim [2π₯ + 1 β π₯ (ππππ‘π (π₯) + 2 ) β 2π₯ ] = lim (1 β π₯ (ππππ‘π (π₯) +
π₯β+β π )) = 2
π₯β+β
π₯β+β
1
Z istnienia powyΕΌszych dwΓ³ch granic wΕaΕciwych wynika istnienie asymptoty ukoΕnej w +β: π¦ = 2π₯ + 1 b.
π(π₯) lim π₯βββ π₯
= lim
π₯βββ
1 π₯
1 π₯
π 2
2π₯+1β (ππππ‘π( )+ ) π₯
= 1
2π₯ lim π₯β+β π₯
+
1 lim π₯β+β π₯
1
π
β lim
π₯β+β
1 π₯ π₯2
ππππ‘π( )+
π 2
=2 1
1
lim [π(π₯) β 2π₯)] = lim [2π₯ + 1 β π₯ (ππππ‘π (π₯) + 2 ) β 2π₯ ] = lim (1 β π₯ (ππππ‘π (π₯) +
π₯βββ π )) = 2
π₯βββ
1
π₯βββ
Taka sama asymptota w ββ. OdpowiedΕΊ: Funkcja π ma dwie asymptoty. JednΔ
pionowΔ
prawostronnΔ
dla π = π oraz jednΔ
ukoΕnΔ
w +β π β β o rΓ³wnaniu: π = ππ + π