可以
当x→0-,arctan 1/x→-π/2
当x→0+,arctan 1/x→+π/2
lim[x→0+] xarctan(1/x)=0×π/2=0
lim[x→0+] xarctan(1/x)=0×(-π/2)=0
故lim[x→0]xarctan(1/x)=0