图中8 9 10 11 怎么解

1个回答

  • (8)当x→-∞时,ln(1+e^x)∽e^x ln(1+2^x)∽2^x

    原式=lim[x→-∞](e^x/2^x)=lim[x→-∞](e/2)^x=0

    (9)属于“∞/∞”型,用罗比塔法则

    原式=lim[x→+∞][e^x/(1+e^x)]/[2^xln2/(1+2^x)]

    =lim[x→+∞][e^x/(1+e^x)]/lim[x→+∞][2^xln2/(1+2^x)]

    =lim[x→+∞][e^x/e^x)]/lim[x→+∞][2^xln2ln2/(2^xln2)]

    =ln2

    (10) 属于“0*∞”型,化为“0/0"用罗比塔法则

    原式=lim[x→+∞][5^(2/x^3)-1]/x^(-3)

    =lim[x→+∞][5^(2/x^3)ln5/[-3x^(-4)=-∞

    (11) 原式=lim[x→0]{e^x[1-e^(tanx-x)]}/(x-tanx)

    =lim[x→+0]{e^x[e^(tanx-x)-1]}/(tanx-x)

    =1