<s>[INST] prove the sign of $1+2x-e^x$ around 0 I want to prove that the sign of $1+2x-e^x$  around $0$ is negative if $x\to 0^-$ and positive if $x\to 0^+$. If I replace $e^x$ by its equivalent function (the tangent at 0) I conclude that the sign of this expression depends on the sign of $x$ as I have $1+2x - 1-x $.
I need this step to compute the limit of $\frac{1+2x+e^x}{1+2x-e^x}$ when $x\to 0$
 [/INST] You should already know from the definition that:
$$
e^x=\lim_{n\to\infty}(1+\frac xn)^n=1+x+\frac{x^2}{2!}+...+\frac{x^k}{k!}+...
$$
Obviously for all $x>0$:   $e^x>1+x$. To consider negative $x$'s we write:
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}(1-\frac x4)+...+\frac{x^{2m+1}}{(2m+1)!}(1-\frac x{2m+2})+...
$$
Thus for small negative $x$, all the terms starting on the $\frac{x^3}{3!}(1-\frac x4) $ term are negative and we conclude:
$$
e^x<1+x+\frac{x^2}2
$$
for small negative $x$.
From the above we get that $1+2x-e^x>1+2x-(1+x)=x>0$ for small positive $x$ and  $1+2x-e^x<1+2x-(1+x+\dfrac{x^2}{2})=x-\dfrac{x^2}{2}<0 $ for small negative $x$.
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