www.fltk.net > 函数y=(x%1)E^(π/2+ArCtAnx)渐近线,B为什么等于.^π.不是等于%E^π?

函数y=(x%1)E^(π/2+ArCtAnx)渐近线,B为什么等于.^π.不是等于%E^π?

k=lim(x→+∞)(x-1)e^(π/2+arctanx)/x=e^π B=lim(x→+∞)[(x-1)e^(π/2+arctanx)-e^πx)]=lim(x→+∞)[(x-1)e^(π/2+arctanx)-e^π(x-1+1))=lim(x→+∞){(x-1)[e^(π/2+arctanx)-e^π)]-e^π} ∵lim(x→+∞)(x-1)[e^(π/2+arctanx)-e^π)]=lim(x→+∞)[e^(π/2+arctanx)-e^π)]

x-->+∞时(π/2)+arctanx-->π,曲线y=(x-1)e^[(π/2)+arctanx]有渐近线y=(x-1)e^π,同理,x-->-∞时(π/2)+arctanx-->0,曲线y=(x-1)e^[(π/2)+arctanx]有渐近线y=(x-1).

求函数y=(x-1)*e^(π/2+arctanx)的斜渐近线 解:x→+∞lim[(x-1)*e^(π/2+arctanx)]/x=x→+∞lime^(π/2+arctanx)-[x→+∞lim[e^(π/2+arctanx)]/x]=e^π=a x→+∞lim[(x-1)*e^(π/2+arctanx)-(e^π)x]=x→+∞lim[e^(π/2+arctanx)-(e^π)]x-{x→+∞lim[-e^(π/2+arctanx)]}=e^π=b

y=(x-1)e^(π/2+arctanx)lim(x→0)(x-1)e^(π/2+arctanx)=-e^(π/2)水平渐近线y=-e^(π/2)(x-1)e^arctanx=ye^(-π/2)lim(y→0)ye^(-π/2)=0lim(x→1)(x-1)e^(arctanx)=0垂直渐近线x=1lim(x→+∞)(x-1)e^(π/2+arctanx)=(x-1)e^πx>0斜渐近线y=(x-1)e^π (x>0)lim(x→-∞)(x-1)e^(π/2+arctanx)=(x-1)x

说明:曲线是y=(x-1)e^(πarctanx/2)吧.若是,求解如下.解:显然,此题没有垂直渐近线,只有斜渐近线 设它的斜渐近线为y=ax+b ∵a=lim(x->±∞)[(x-1)e^(πarctanx/2)/x] =lim(x->±∞)[(1-1/x)e^(πarctanx/2)] =(1-0)e^((π/2)(π/2)) =e^(±π/4) b=lim(x->±∞)

x=0,y=0,图象一支过坐标原点;仅考虑反三角函数取值在-л/2arctanxл/2的情况(其它处类似);lim(y/x)=lim[arctan(x)]=±л/2,渐近线斜率为±л/2;或由y'=arctan(x)+x/(1+

求得直线的斜率k后再求截矩b.b=y-kx取极限.以当x趋于负无穷时为例,b=lim(x-1)e^(pi/2+arctanx)-x=limx(e^(pi/2+arctanx)-1)-lime^(pi/2+arctanx)=-1-1=-2( 前一个极限可以用洛必达法则做).

没有水平渐近线,斜渐近线倒是有两条:y=x+π/2和y=x-π/2.x趋于无穷时,y也趋于无穷,所以没有水平渐近线.如果x趋于正无穷时,y趋于某个常数C1,那么有水平渐近线y=C1;如果x趋于负无穷时,y趋于某个常数C2,那么有水平渐

∵limx→∞e1x2arctanx2+x+1(x-1)(x+2)=1arctan1=π4∴y=π4是其水平渐近线∵limx→0e1x2arctanx2+x+1(x-1)(x+2)=∞∴x=0是其铅直渐近线又limx→∞yx=limx→∞e1x2xarctanx2+x+1(x-1)(x+2)=0∴没有斜渐近线.∴有两条渐近线故选:B.

y=(x-1)e^(π/2+arctanx)lim(x→0)(x-1)e^(π/2+arctanx)=-e^(π/2)水平渐近线y=-e^(π/2)(x-1)e^arctanx=ye^(-π/2)lim(y→0)ye^(-π/2)=0lim(x→1)(x-1)e^(arctanx)=0垂直渐近线x=1y=(x-1)*e^(π/2+arctanx)的斜渐近线x→+∞lim[(x-1)*e^(π/2+arctanx)]/x=x→+∞lime^(

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