Oliver 發問時間: 科學數學 · 10 年前

極限的8個題目求解

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您好,剛剛核對答案,發現有幾題不一樣,(5)-1/2 (13)0 (14)-1 (18)A=-3/√2,B=1,a=-3,b=-6 另外可以請你幫我說明一下第20題嗎?這一步(1+Cosx/x)/(1+1/x)如何得到答案的呢?

1 個解答

評分
  • 10 年前
    最佳解答

    1 lim(x->n)_[x-[x-1]]

    Sol

    lim(x->n-)_[x-[x-1]]

    = lim(x->n-)_[x-n+2]

    =(n-1)-n+2

    =1

    lim(x->n+)_[x-[x-1]]

    = lim(x->n-)_[x-n+1]

    =n-n+1

    =1

    So

    lim(x->n)_[x-[x-1]]=1

    5. lim(x->∞)_x√(-x)/(√(1-4x^3)

    =lim(x->∞)_-1/√x/(√(1/x^3-1)

    =1

    6. lim(x->∞)_√x[(√(x+2)-√x]

    =lim(x->∞)_√x[√(x+2)-√x]*[√(x+2)+√x]/[√(x+2)+√x]

    =lim(x->∞)_√x[(x+2)-x]/[√(x+2)+√x]

    =lim(x->∞)_2/[√(1+2/x)+√1]

    =1

    7. lim(x->∞)_[√(2x^2+1)-√(x^2-x)]

    =lim(x->∞)_x[√(2+1/x^2)-√(1-1/x)]

    =∞

    13. lim(x->n-)_{[x]-x}= lim(x->n-)_{n-1-x}=-1

    14. lim(x->n+)_{[x]-x}= lim(x->n+)_{n-x}=0

    lim(x->n-)_{[x]-x}= lim(x->n-)_{n-1-x}=-1

    極限不存在

    18.

    (1) A=lim(t->-∞)_(2t+4)/√(2t^2-5)

    =lim(t->-∞)_(2+4/t)/√(2-5/t^2)

    =√2

    B=lim(t->∞)_Sin(1/t)/Tan(1/t)=Sin0/Tan0=1

    (2) lim(x->-3+)_(ax+2b)=-3a+2b=(-2√2/3)*√2*(-3)-a=8-a

    -2a+2b= 8

    a-b=4--------------------------------

    lim(x->3-)_(b-5*1*3)= 3a +2b

    b-8= 3a +2b

    -3a-b=8

    3a +b=-8----------------------------

    a=-1,b=-5

    20.lim(x->∞)_(x+Cosx)/(x+1)

    = lim(x->∞)_(1+Cosx/x)/(1+1/x)

    =1

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