martes, 22 de noviembre de 2011

COLEGIO DE ESTUDIOS CIENTIFICOS Y TECNOLOGICOS DEL ESTADO DE OAXACA

TRABAJO:   DERIVADAS Y LOGARITMOS “EJERCICIOS”


MATERIA: CALCULO DIFERENCIAL


PROFESOR: ING. JOEL ALEGRIA SALINAS


INTEGRANTES DEL EQUIPO:   ABISAY VELASCO CHAVEZ
                                  ISMAEL GOMEZ YESCAS
                                          RUBICELIA YESCAS ANGELES
                                                 MIGUEL A. HERNANDEZ JERONIMO
                                 ADRIAN TORO MENDEZ
                                         MIRIAM LOPEZ BERNARDINO
                                         ALDO R. VASQUEZ JERONIMO

                                                              SEMESTRE:      5                           GRUPO: 501


SAN MIGUEL TALEA DE CASTRO A 23 DE NOVIEMBRE DE 2011.










miércoles, 7 de septiembre de 2011

LIMITES(PROBLEMAS)

Ejercicios de límite
                〖lim〗┬(X→∞)⁡〖(〖 2x〗^3+3x-2)/(4x^2+2x-3)〗

lim┬(X→∞)⁡〖(〖 2x〗^3+3x-2)/(4x^2+2x-3)=(〖2(∞)〗^3+3(∞)-2)/(4〖(∞)〗^2+2(∞)-3)=(2(∞)+3(∞)-2)/(4(∞)+2(∞)-3)=(∞+∞-2)/(∞+∞-3)=∞/∞〗=∞


lim┬(X→∞)⁡〖(〖 2x〗^3/〖 x〗^3 +3x/〖 x〗^3 -2/〖 x〗^3 )/((4x^2)/〖 x〗^3 +2x/〖 x〗^3 -3/〖 x〗^3 )=(2+3/x^2 -2/〖 x〗^3 )/(4+2/x^2 -3/〖 x〗^3 )=(2+3/∞-2/∞)/(4+2/∞-3/∞)=(2+0+0)/(4+0+0)=2/4〗

                〖lim〗┬(X→∞)⁡〖(x^2+x-5)/(4x^2+7)〗


lim┬(X→∞)⁡〖(x^2+x-5)/(4x^2+7)=(〖(∞)〗^2+∞+5)/(4〖(∞)〗^2+7)=(∞+∞+5)/(4(∞)+7)=(∞+∞+5)/(∞+7)=∞/∞〗=∞

lim┬(X→∞)⁡〖(〖 x〗^2/〖 x〗^2 +x/〖 x〗^2 +5/〖 x〗^2 )/((4x^2)/〖 x〗^² +7/〖 x〗^2 )=(1+x+5/〖 x〗^2 )/(4+7/x^2 )=(1+∞+5/∞)/(4+7/∞)=(1+0+0)/(4+0)=1/4〗



                〖lim〗┬(X→∞)⁡〖7/(x+3)〗

lim┬(X→∞)⁡〖7/(x+3)=7/(∞+3)=7/∞=0〗



                〖lim〗┬(X→∞)⁡〖(〖 4x〗^4+5x-2)/(3x^4+x^3-x)〗

lim┬(X→∞)⁡〖(〖 4x〗^2+5x-2)/(3x^4+x³-x)=(〖4(∞)〗^2+5(∞)-2)/(4〖(∞)〗^4+(∞)3-∞)=(4(∞)+5(∞)-2)/(3(∞)+(∞)-∞)=(∞+∞-2)/∞=∞/∞〗=0
lim┬(X→∞)⁡〖(〖 4x〗^2/〖 x〗^4 +5x/〖 x〗^4 -2/〖 x〗^4 )/((3x^4)/〖 x〗^4 +x³/〖 x〗^4 -x/〖 x〗^4 -)=(4/〖 x〗^2 +5/〖 x〗^(3   ) -2/〖 x〗^4 )/(3+x-x³)=(4/∞+5/∞-2/∞)/(3+∞-∞)=(0+0+0)/(3+∞-∞)=0〗

                〖lim〗┬(X→∞)⁡〖(〖 6x〗^3+4x-3)/(2x^3-3x+5)〗

lim┬(X→∞)⁡〖(〖 6x〗^3+4x-3)/(2x^3-3x+5)=(〖6(∞)〗^3+4(∞)-3)/(2〖(∞)〗^3-3(∞)+5)=(6(∞)+4(∞)-3)/(2(∞)-3(∞)+5)=(∞+∞-3)/(∞-∞+5)=∞/∞〗=∞

lim┬(X→∞)⁡〖(〖 6x〗^3/〖 x〗^3 +4x/〖 x〗^3 -3/〖 x〗^3 )/((2x^3)/〖 x〗^3 -3x/〖 x〗^3 -5/〖 x〗^3 )=(6+4/x^2 -3/〖 x〗^3 )/(2-3/x^2 +5/〖 x〗^3 )=(6+4/∞-3/∞)/(2-3/∞+5/∞)=(6+0-0)/(2-0+0)=6/2〗




                〖lim〗┬(X→∞)⁡〖〖 7x〗^3-8x+5〗

lim┬(X→∞)⁡〖7x³-8x+5=7(∞)-8(∞)+5=7(∞)-8(∞)+5=∞-∞+5=0〗


lim┬(X→∞)⁡〖(〖 7x〗^3/〖 x〗^3 -8x/〖 x〗^3 +5/〖 x〗^3 )/=(7-8/x^2 +5/〖 x〗^3 )/=(7-8/∞+5/∞)/=(7-0+0)/=7〗



                〖lim〗┬(X→∞)⁡〖5/x〗

lim┬(X→∞)⁡〖5/x=5/∞=0=〗



lim┬(X→∞)⁡〖5/x=5x/1〗=5(∞)/1=((∞))/1=∞




                〖lim〗┬(X→∞)⁡〖〖 7x〗^3-8x+5〗

lim┬(X→∞)⁡〖5x²-2x+3=5(∞)²-2(∞)+3=5(∞)-2(∞)+3=∞-∞+3=3〗

5x²/x²-2x/x²+3/x²=5-2/x+3/x²=5-2/∞+5/∞=0



                〖lim〗┬(X→-1)⁡〖(〖 x〗^2-2x-3)/(x^2+3x+2)〗

lim┬(X→-1)⁡〖(〖 x〗^2-2x-3)/(x^2+3x+2)=(〖(-1)〗^2-2(-1)-3)/(〖(-1)〗^2+3(-1)+2)=(1+2-3)/(1-3+2)=0/0〗=0

lim┬(X→-1)⁡〖((x+1)(x-3))/((x+1)(x+2))=(-1-3)/((-1)+2)=(-4)/1=-4〗



                〖lim〗┬(X→∞)⁡〖(〖 x〗^2-3x-2)/(x^2-5x+6)〗

lim┬(X→-1)⁡〖(〖 x〗^2-3x-2)/(x^2-5x+6)=(〖(2)〗^2-3(1)-2)/(〖(2)〗^2-5(2)+6)=(4-6-2)/(4-10+6)=(-4)/0〗=E.M



                〖lim〗┬(X→∞)⁡〖4x²-3x-1〗


lim┬(X→∞)⁡〖4x²-3x-1=4(∞)²-3(∞)-1=4(∞)-3(∞)-1=∞-∞-1=-1〗


lim┬(X→-1)⁡〖4x²/x²-3x/x²-(-1)/x²=4-3/x-1/x²=4-3/∞-1/∞=4-0-0=4〗





                〖lim〗┬(X→∞)⁡〖(〖 2x〗^3+3x-2)/(4x^2+2x-3)〗

lim┬(X→∞)⁡〖(〖 2x〗^3+3x-2)/(4x^2+2x-3)=(〖2(∞)〗^3+3(∞)-2)/(4〖(∞)〗^2+2(∞)-3)=(2(∞)+3(∞)-2)/(4(∞)+2(∞)-3)=(∞+∞-2)/(∞+∞-3)=∞/∞〗=∞
lim┬(X→-1)⁡〖(-3x²)/x²+5x/x²-7/x²=-3+5/x-7/x²=-3+5/∞-7/∞=-3+0-0=-3〗

                〖lim〗┬(X→4)⁡〖(〖 x〗^2-x-12)/(x^2+x-20)〗

lim┬(X→4)⁡〖(〖 x〗^2-x-12)/(x^2+x-20)=(〖(4)〗^2-4-12)/(〖(4)〗^2+4-20)=(16-4-12)/(16+4-2)=0/0〗=0

lim┬(X→4)⁡〖((x-4)(x+3))/((x+4)(x+5))=(4+3)/(4+5)=7/9=-4〗


                〖lim〗┬(X→-2)⁡〖(〖 x〗^2+7x+10)/(x^2-x-6)〗


lim┬(X→-1)⁡〖(〖 x〗^2+7x+10)/(x^2-x-6)=(〖(-2)〗^2+2(-2)+10)/(〖(-2)〗^2-(-2)-6)=(4-14+10)/(4+2-6)=0/0〗=0

lim┬(X→-1)⁡〖((x+5)(x+2))/((x-3)(x+2))=(-2+5)/((-2)-3)=3/(-5)=-〗  (-3)/5



                〖lim〗┬(X→2)⁡〖(〖 x〗^2+x-6)/(x^2-6)〗


lim┬(X→-1)⁡〖(〖 x〗^2+x-6)/(x^2-6)=(〖(2)〗^2+2-6)/((2)^2-4)=(4+2-6)/(4-4)=0/0〗=0

lim┬(X→-1)⁡〖((x+3)(x-2))/((x-2)(x-2))=(2+3)/(2+2)=5/4〗


                〖lim〗┬(X→4)⁡〖(〖 x〗^2-x-12)/(x-4)〗


lim┬(X→4)⁡〖(〖 x〗^2-x-12)/(x-4)=(〖(4)〗^2-4-12)/(4-4)=(16-4-12)/(4-4)=0/0〗=0


lim┬(X→4)⁡〖((x-4)(x+3))/((x-4)(1+0))=(4+3)/1=7/1=7〗