limx→∞(1+1x)x=
e
limx→∞(1+1x)x
At x=0, the value of the given expression is 1∞ form.
We can re-write as
limx→∞exloge(1+1x)
Expansion of loge(1+x)=x−x22+x33.......
⇒limx→∞ex[1x−12x2+13x3−.....]
=limx→∞e[1−12x+13x3−.....]
put x=∞
We have e[1−0....]
=e