 \lim_{ x \to 0 }\frac{e^x-x-1}{x²}=\frac{1}{2}

1+x+\frac{x²}{2}≤e^x≤1+x+\frac{x²}{2}+x^3   \;   \;  ,  \;  x\in[0,1]

1+x+\frac{x²}{2}+x^3≤e^x≤1+x+\frac{x²}{2}   \;   \;  ,  \;  x\in[-1,0]

$ ∀x∈ℝ  \;   \;   \; , \;   \;  1+x≤ e^x $   

g(x) = ℯ^(x)
f(x) = (ℯ^(x) - 1) / x
d(x) = 1 / 3 * x^(3) + 1 / 2 * x^(2) + x + 1