大一高等数学求极限1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限2 cotx -1/x当x趋于0的极限3

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  • 1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限

    =lim(x->π/2)1/(x-π/2)/sec²x

    =lim(x->π/2)cos²x/(x-π/2)

    =lim(x->π/2)2cosx(-sinx)/1

    =0

    2.

    lim(x->0)cotx-1/x

    =lim(x->0)xosx/sinx-1/x

    =lim(x->0)(xcosx-sinx)/xsinx

    =lim(x->0)(xcosx-sinx)/x²

    =lim(x->0)(cosx-xsinx-cosx)/2x

    =lim(x->0)(-xsinx)/2x

    =lim(x->0)(-sinx)/2

    =0

    3.(x^3+x^2+x+1)^1/3-x

    =lim(x->∞)e^ln(x^3+x^2+x+1)^1/3-x

    =lim(x->∞)e^[ln(x^3+x^2+x+1)]/(3-x)

    =e^lim(x->∞)1/(x^3+x^2+x+1) *(3x²+2x+1)/(-1)

    =e^lim(x->∞)-(3x²+2x+1)/(x^3+x^2+x+1)

    =e^lim(x->∞)-(6x+2)/(3x^2+2x+1)

    =e^lim(x->∞)-(6)/(6x+2)

    =e^0

    =1