limx→0 xe1+x2+x4−1x−11+x2+x4−1
does not exist
is equal to 1
is equal
is equal to 0
We find that
limx→0 1+x2+x4−1x=limx→0 x21+x2x1+x2+x4+1=0
Let y=1+x2+x4−1x.Then,
∴ limx→0 xe1+x2+x4−1x−11+x2+x4−1
=limx→0 e1+x2+x4−1x−11+x2+x4−1x=limy→0 ey−1y=1