evaluate lim_x-0 (e^x-sinx-1)/x

limx→∞(1+1x)x=e (number of Neper), and also this limit:

limx→0(1+x)1x=e that it is easy to demonstrate in this way:

let x=1t, so when x→0 than t→∞ and this limit becomes the first one.

So:

let ex−1=t⇒ex=t+1⇒x=ln(t+1)

and if x→0⇒t→0

limx→0 ex−1 X=limt→0 tln(t+1)=limt→0 1ln(t+1)t=

=limt→0 11tln(t+1)=limt→0 1ln(t+1)1t=

(for the second limit)=1lne=11=1.