Question

__________________??

Answer 1

limx→0ex−1x=1(1)

because

limx→0ex−1x=limx→0[(1+x)1x]x−1x=limx→01+x−1x=1

Now we want to make use of this fact to calculate the required limit. Because of (1) we have the following:

limx→0e2x−2ex+1x2=1(2)

Let the required limit be L. Rewrite the limit in the following form:

L=lim2x→0e2x−1−2x(2x)2

So multiplying both sides by 4, we have

4L=limx→0e2x−1−2xx2(3)

Hope you see where I'm getting. I want to rewrite L in a way that we can use (2).

Now consider the following:

limx→0e2x−2ex+1x21L=limx→0e2x−1−2x−2ex+2+2xx2=4L−2limx→0ex−1−xx2=4L−2L=12

I think you can generalize this to calculate like

limx→0ex−1−x−x22x3