Simple analytic proof. If asked to prove that $$e^x>1+x: x>0$$
Can I argue that
$$\lim_{x\rightarrow0}\frac{e^x-1}{x}=1$$
and this limit is approached from right side. However, am not confident how I justify it approaches from right side or whether that justification suffices.
Thank You. 
 A: By the Mean Value Theorem, 
$$\frac{e^x-e^0}{x-0}=e^{\xi}$$
for some $\xi$ between $0$ and $x$. Thus $e^{\xi}\gt 1$. 
A: HINT: Use series expansion of $e^x$

$e^x=1+x+\frac{x^2}{2!}+\cdots >1+x$
since $\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ is positive quantity.

Alternative:
As given in julien's comment just study the function $f(x)=e^x-x-1$.Note that $f(0)=0$ and
take derivative of $f(x)$, you get $f'(x)=e^x-1$. It is strictly greater than $0$. Otherwise $f''(x)=e^x$. Clearly $f''(x)>0$ since $e^x$ is a non-negative function.Here you can conclude $f(x)>0 \implies e^x >1+x$ .
A: For $f(x)=e^x-1-x$ we have
$$f(0)=0,$$
and for all $x>0$
$$f'(x)=e^x-1>0 \;\;  \Rightarrow \;\;f(x)>0.
$$
A: If you know what a derivative is then take $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=e^x-x-1$ and then $f^{'}(x)=e^x-1$ so $f^{'}(x)=0 $ iff $x=0$.
Now for $x>0$ => $f^{'}(x)>0$ and for $x<0$ => $f^{'}(x)<0$. 
From $x>0$ => $f^{'}(x)>0$ you get that the function $f$ increases on the positive real numbers so for these values ($x>0$)
$f(x)>f(0)$ which means $e^x-x-1>0$ ,$\forall x\in (0,\infty)$.
