简单的极限题。。。
利用等价无穷小的性质,求下列极限。1.lim〔(sinx^n)〕/(tanx)^m2.lim(tanx-sinx)/sin^3(x)两者都是x→0...
利用等价无穷小的性质,求下列极限。1.lim〔(sinx^n)〕/(tanx)^m 2.lim(tanx-sinx) /sin^3(x) 两者都是x→0
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1.lim(x→0)〔(sinx^n)〕/(tanx)^m(运用等价无穷小代换)
=lim(x→0)x^n/x^m
=lim(x→0)x^(n-m)
若n>m则极限为0
若n=m则极限为1
若n<m则极限未知
lim(x→0)(tanx-sinx) /sin^3(x)(运用等价无穷小代换)
=lim(x→0)(tanx-sinx)/x^3 (这是0/0型,运用洛必达法则)
=lim(x→0)(sec^2x-cosx)/(3x^2)
=lim(x→0)(1/cos^2x-cosx)/(3x^2)
=lim(x→0)(1-cos^3x)/(3cos^2x*x^2)
=lim(x→0)(1-cos^3x)/(3x^2) (这是0/0型,运用洛必达法则)
=lim(x→0)(3cos^2xsinx)/(6x)
=1/2
=lim(x→0)x^n/x^m
=lim(x→0)x^(n-m)
若n>m则极限为0
若n=m则极限为1
若n<m则极限未知
lim(x→0)(tanx-sinx) /sin^3(x)(运用等价无穷小代换)
=lim(x→0)(tanx-sinx)/x^3 (这是0/0型,运用洛必达法则)
=lim(x→0)(sec^2x-cosx)/(3x^2)
=lim(x→0)(1/cos^2x-cosx)/(3x^2)
=lim(x→0)(1-cos^3x)/(3cos^2x*x^2)
=lim(x→0)(1-cos^3x)/(3x^2) (这是0/0型,运用洛必达法则)
=lim(x→0)(3cos^2xsinx)/(6x)
=1/2
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1.lim(x→0) sin(x^n) / (tanx)^m = lim(x→0) x^n / x^m
若n>m 则极限为0; 若n=m 则极限为1; 若n<m 则极限 = ∞
2. lim(x→0)((anx-sinx) /sin³x = lim(x→0) (tanx-sinx) / x³
= lim(x→0) tanx( 1 - cosx) / x³ 1 - cosx ~ x²/2
= lim(x→0) x * x²/2 / x³
= 1/2
若n>m 则极限为0; 若n=m 则极限为1; 若n<m 则极限 = ∞
2. lim(x→0)((anx-sinx) /sin³x = lim(x→0) (tanx-sinx) / x³
= lim(x→0) tanx( 1 - cosx) / x³ 1 - cosx ~ x²/2
= lim(x→0) x * x²/2 / x³
= 1/2
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