设f(x，y)＝y1+xy−1−ysinπxyarctanx，x＞0，y＞0，求（Ⅰ）g(x)＝limy→+∞f(x，y)；（Ⅱ）limx→0+g(x)．

Question

设f(x，y)＝(y)/(1+xy)−(1−ysin(πx)/(y))/(arctanx)，x＞0，y＞0，求
（Ⅰ）g(x)＝

lim

y→+∞

f(x，y)；
（Ⅱ）

lim

x→0+

g(x)．

Answer 1

（Ⅰ） g(x)＝

lim

y→+∞

f(x，y)＝

lim

y→∞

((y)/(1+xy)−(1−ysin(πx)/(y))/(arctanx))=

lim

y→∞

((1)/((1)/(y)+x)−(1−(sin(πx)/(y))/((1)/(y)))/(arctanx))＝(1)/(x)−(1−πx)/(arctanx)．
（Ⅱ） 

lim

x→0+

g(x)＝

lim

x→0+

((1)/(x)−(1−πx)/(arctanx))＝

lim

x→0+

(arctanx−x+πx2)/(xarctanx)（通分）=

lim

x→0+

(arctanx−x+πx2)/(x2)＝

lim

x→0+

((1)/(1+x2)−1+2πx)/(2x)=

lim

x→0+

(−x2+2πx(1+x2))/(2x)＝π．