lim(x→0,y→0) xy/(√2-e^xy)-1=?如题

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/24 18:43:14
lim(x→0,y→0)xy/(√2-e^xy)-1=?如题lim(x→0,y→0)xy/(√2-e^xy)-1=?如题lim(x→0,y→0)xy/(√2-e^xy)-1=?如题应该是:lim(x-

lim(x→0,y→0) xy/(√2-e^xy)-1=?如题
lim(x→0,y→0) xy/(√2-e^xy)-1=?
如题

lim(x→0,y→0) xy/(√2-e^xy)-1=?如题
应该是:lim(x->0,y->0) xy/[√(2-e^xy)-1]
这是0/0型极限式,用二元函数极限的洛必达法则公式:
lim(x->x0,y->y0) [f(x,y)/g(x,y)]
=lim(x->x0,y->y0) {[f'x(x,y)dx+f'y(x,y)dy]/[g'x(x,y)dx+g'y(x,y)dy]}
其中,dx=x-x0,dy=y-y0;
对于本题,f(x,y)=xy,g(x,y)=√(2-e^xy)-1
f'x(x,y)=y,f'y(x,y)=x;
g'x(x,y)=(-e^xy)*y*1/2*1/√(2-e^xy),g'y(x,y)=(-e^xy)*x*1/2*1/√(2-e^xy)
dx=x-0=x,dy=y-0=y
∴lim(x->0,y->0) xy/[√(2-e^xy)-1]
=lim(x->0,y->0) [ydx+xdy]*2√(2-e^xy)/[(-e^xy)*(ydx+xdy)]
=lim(x->0,y->0) √(2-e^xy)/(-e^xy)
=√(2-e^0)/(-e^0)
=-1