Original formula = lim [x-> 0+] [e (xlnsinx)-1/(CHX) (2/x 2]]

=e^lim[x->； 0][LNS inx+xcos x/sinx]-lim[x-& gt； 0+] 1/[(chx)^2]^( 1/x^2)

= 0- 1/lim[x->； 0+]{[ 1+(shx)^2]^[ 1/(shx)^2]}^[(shx)^2/x^2]

=- 1/e

Doesn't match your answer. It seems that the topic is wrong, so change it to the topic you modified:

Original formula = lim [x-> 0+] [sinx/x-1/(CHX) (2/x 2]]

= 1-lim[x->； 0+] 1/[(chx)^2]^( 1/x^2)

= 1- 1/lim[x->； 0+]{[ 1+(shx)^2]^[ 1/(shx)^2]}^[(shx)^2/x^2]

= 1- 1/e

Choose B.

I made two questions. The first one was submitted to encourage adoption.