用洛必达法则求下列极限
请写详细步骤或着解题思路 1)(x趋近0右)lim[(lntan3x)/(lntan4x)] 2) (x趋近0)lim x cot3x 3) (x趋近0右)lim (x^tanx) 4) (x趋近pi/2)lim[tanx/(tan5x)]
详细解答如下:
1、lim(x→0+) ln(tan3x)/ln(tan4x) =lim(x→0+) 1/tan3x×(sec3x)^2)×3/(1/tan4x×(sec4x)^2×4) =lim(x→0+) tan4x/tan3x×(cos4x)^2/(cos3x)^2×3/4 =3/4×lim(x→0+) tan4x/tan3x =3/4×lim(x→0+) 4x/3x(等价无穷小替换) =3/4×4/3=1 2、lim(x→0) x×cotx =lim(x→0) x/tanx =lim(x→0) 1/(secx)^2 =lim(x→0) (cosx)^2=1 3、lim(x→0) x^(tanx)=lim(x→0) e^(tanx×lnx) lim(x→0) tanx×lnx =lim(x→0) x×lnx (等价无穷小替换) =lim(x→0) lnx/(1/x) =lim(x→0) 1/x / (-1/x^2) =lim(x→0) (-x)=0 所以,lim(x→0) x^(tanx)=lim(x→0) e^(tanx×lnx)=e^0=1 4、lim(x→π/2) tanx/tan5x =lim(x→π/2) sinx/sin5x×cos5x/cosx =sin(π/2)/sin(5π/2)×lim(x→π/2) cos5x/cosx =lim(x→π/2) cos5x/cosx =lim(x→π/2) -sin5x×5/(-sinx) =5×sin(5π/2)/sin(π/2)=5。
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