Question

这道题能用洛必达法则做吗，写下详细步骤，谢谢，满意采纳

函数极限

Answer 1

y=1/x
lim(x->∞) x. sin[ 2x/(x^2+1) ]
=lim(y->0) sin[ (2/y)/(1/y^2+1) ] / y
=lim(y->0) sin[ 2y/(y^2+1) ] / y
=lim(y->0) [ 2y/(y^2+1) ] / y
=lim(y->0) 2/(y^2+1)
=2/(1+0)
=2