Define the natural number e by e=limx→0(1+x)1/x.
Then, I can prove limx→0ex−1x=1.
Let z=ex−1. Then, x=ln(z+1) and limx→0ex−1x=limz→0zln(z+1)=1lne=1.
Using a similar trick (without L'Hoptial's rule), can we prove limx→0ex−1−xx2=12?
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
No comments:
Post a Comment