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Задачи од граница на функција со Лопиталово правило

Module by: Beti Andonovic. E-mail the author

Summary: Solved exercises on limits of functions using L'Hopital's rule Решени задачи од лимеси со Лопиталово правило

Лимеси со Лопиталово правило

1. limx1exex1=ЛПlimx1(exe)'(x1)'=limx1ex=e1=elimx1exex1=ЛПlimx1(exe)'(x1)'=limx1ex=e1=e size 12{ {"lim"} cSub { size 8{x rightarrow 1} } { {e rSup { size 8{x} } - e} over {x - 1} } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 1} } { { \( e rSup { size 8{x} } - e \) '} over { \( x - 1 \) '} } = {"lim"} cSub { size 8{x rightarrow 1} } e rSup { size 8{x} } =e rSup { size 8{1} } =e} {}

2. limx0ex2cosxx2=ЛПlimx0(ex2cosx)'(x2)'=limx0ex22x+sinx2x=ЛПlimx0ex2cosxx2=ЛПlimx0(ex2cosx)'(x2)'=limx0ex22x+sinx2x=ЛП size 12{ {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{x rSup { size 6{2} } } } - "cos"x} over {x rSup {2} } } { size 24{ {}={}} } cSup {"ЛП"} { size 12{"lim"} } cSub {x rightarrow 0} { { size 12{ \( e rSup {x rSup { size 6{2} } } size 12{ - "cos"x \) '}} } over { size 12{ \( x rSup {2} size 12{ \) '}} } } size 12{ {}= {"lim"} cSub {x rightarrow 0} { { size 12{e rSup {x rSup { size 6{2} } } size 12{ cdot 2x+"sin"x}} } over { size 12{2x} } } { size 24{ {}={}} } cSup {"ЛП"} }} {}

= ЛП lim x 0 ( e x 2 2x + sin x ) ' ( 2x ) ' = lim x 0 e x 2 2x + 2 e x + cos x 2 = = ЛП lim x 0 ( e x 2 2x + sin x ) ' ( 2x ) ' = lim x 0 e x 2 2x + 2 e x + cos x 2 = size 12{ { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { \( e rSup { size 8{x rSup { size 6{2} } } } cdot 2x+"sin"x \) '} over { \( 2x \) '} } = {"lim"} cSub {x rightarrow 0} { { size 12{e rSup {x rSup { size 6{2} } } size 12{ cdot 2x+2e rSup {x} } size 12{+"cos"x}} } over { size 12{2} } } size 12{ {}={}}} {}
(1)
= 1 0 + 2 1 + 1 2 = 3 2 = 1 0 + 2 1 + 1 2 = 3 2 size 12{ {}= { {1 cdot 0+2 cdot 1+1} over {2} } = { {3} over {2} } } {}
(2)

3. limx0exexsinx=ЛПlimx0(exex)'(sinx)'=limx0ex+excosx=1+11=21=2limx0exexsinx=ЛПlimx0(exex)'(sinx)'=limx0ex+excosx=1+11=21=2 size 12{ {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{x} } - e rSup { size 8{ - x} } } over {"sin"x} } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { \( e rSup { size 8{x} } - e rSup { size 8{ - x} } \) '} over { \( "sin"x \) '} } = {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{x} } +e rSup { size 8{x} } } over {"cos"x} } = { {1+1} over {1} } = { {2} over {1} } =2} {}

4. limx0esin2xesinxx=ЛПlimx0(esin2xesinx)'(x)'=limx0esin2x2cos2xesinxcosx1=limx0esin2xesinxx=ЛПlimx0(esin2xesinx)'(x)'=limx0esin2x2cos2xesinxcosx1= size 12{ {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{"sin"2x} } - e rSup { size 8{"sin"x} } } over {x} } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { \( e rSup { size 8{"sin"2x} } - e rSup { size 8{"sin"x} } \) '} over { \( x \) '} } = {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{"sin"2x} } cdot 2"cos"2x - e rSup { size 8{"sin"x} } cdot "cos"x} over {1} } ={}} {}

= lim x 0 e sin x ( 2 cos 2x cos x ) 1 = 1 2 1 1 1 = 1 1 = 1 = lim x 0 e sin x ( 2 cos 2x cos x ) 1 = 1 2 1 1 1 = 1 1 = 1 size 12{ {}= {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{"sin"x} } \( 2"cos"2x - "cos"x \) } over {1} } = { {1 cdot 2 - 1 cdot 1} over {1} } = { {1} over {1} } =1} {}
(3)

5. limx0eaxebxx=ЛПlimx0(eaxebx)'(x)'=limx0eaxaebxb1=1a1b1=ablimx0eaxebxx=ЛПlimx0(eaxebx)'(x)'=limx0eaxaebxb1=1a1b1=ab size 12{ {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{ ital "ax"} } - e rSup { size 8{ ital "bx"} } } over {x} } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { \( e rSup { size 8{ ital "ax"} } - e rSup { size 8{ ital "bx"} } \) '} over { \( x \) '} } = {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{ ital "ax"} } cdot a - e rSup { size 8{ ital "bx"} } cdot b} over {1} } = { {1 cdot a - 1 cdot b} over {1} } =a - b} {}

6. limx0x(e1x1)=limx0e1x11x=ЛПlimx0(e1x1)'(1x)'limx0e1x(1x2)(1x2)=limx0e1x=e1=e0=1limx0x(e1x1)=limx0e1x11x=ЛПlimx0(e1x1)'(1x)'limx0e1x(1x2)(1x2)=limx0e1x=e1=e0=1 size 12{ {"lim"} cSub { size 8{x rightarrow 0} } x \( e rSup { size 8{ { {1} over {x} } } } - 1 \) = {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{ { {1} over {x} } } } - 1} over { { {1} over {x} } } } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { \( e rSup { size 8{ { {1} over {x} } } } - 1 \) '} over { \( { {1} over {x} } \) '} } {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{ { {1} over {x} } } } cdot \( - { {1} over {x rSup { size 8{2} } } } \) } over { \( - { {1} over {x rSup { size 8{2} } } } \) } } = {"lim"} cSub { size 8{x rightarrow 0} } e rSup { size 8{ { {1} over {x} } } } =e rSup { size 8{ { {1} over { infinity } } } } =e rSup { size 8{0} } =1} {}

7. limx0e2x13x=ЛПlimx0e2x1'3x'=limx0e2x23=2e03=213=23limx0e2x13x=ЛПlimx0e2x1'3x'=limx0e2x23=2e03=213=23 size 12{ {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{2x} } - 1} over {3x} } { size 24{ {}={}} } cSup { size 8{"ЛП"} } {"lim"} cSub { size 8{x rightarrow 0} } { { left (e rSup { size 8{2x} } - 1 right ) rSup { size 8{'} } } over { left (3x right ) rSup { size 8{'} } } } = {"lim"} cSub { size 8{x rightarrow 0} } { {e rSup { size 8{2x} } cdot 2} over {3} } = { {2 cdot e rSup { size 8{0} } } over {3} } = { {2 cdot 1} over {3} } = { {2} over {3} } } {}.

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