设y=y(x)是由方程ln(x^2+y+1)=x^3+sinx确定,求n趋于无穷时n*y(2/n)的极限

Question

设y=y(x)是由方程ln(x^2+y+1)=x^3+sinx确定,求n趋于无穷时n*y(2/n)的极限

Answer 1

ln(x^2+y+1)=x^3+sinx
x^2+y+1=e^(x^3+sinx)
y=e^(x^3+sinx)-x^2-1
y(2/n)=e^(8/n^3+sim(2/n))-4/n^2-1
∴ny(2/n)=n*e^(8/n^3+sin(2/n))-4/n-n
又 n-->∞时,
e^(8/n^3+sin(2/n))~8/n^3+sin(2/n)+1
∴lim(n-->∞)n*Y(2/n)
=lim(n-->∞)n*e^(8/n^3+sin(2/n))-4/n-n
=lim(n-->∞)n[8/n^3+sin(2/n)+1]-4/n-n
=lim(n-->∞)8/n^2+nsin(2/n)+n-4/n-n
=lim(n-->∞)8/n^2+2*sin(2/n)/(2/n)-4/n
=2