Solving $\lim_{x\to0} \frac{e^x-1-x}{x^2}$ $$ = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}$$ 
$$ = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}$$ 
$$ = 1 * \lim_{x\to0} \frac{1}{x} -\lim_{x\to0}\frac{1}{x}$$
$$ = \infty -\infty$$
How do I actually solve this without using L'Hospital. I'm just looking for a hint, not a concrete answer.
 A: No series or L'Hospital's required!  "DISCLAIMER": This is not a hint.
Consider $x\mapsto-x$ to get
$$L=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{-x}-1+x}{x^2}$$
Add these together to get
$$\begin{align}2L&=\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}\tag1\\&=\lim_{x\to0}\frac{(e^{x/2}-e^{-x/2})^2}{x^2}\tag2\\&=\left[\lim_{x\to0}\frac{e^{x/2}-e^{-x/2}}x\right]^2\tag3\\&=\left[\lim_{x\to0}\frac{e^x-e^{-x}}{2x}\right]^2\tag4\\&=\frac14\left[\lim_{x\to0}\frac{e^x-1}x-\lim_{x\to0}\frac{e^{-x}-1}x\right]^2\tag5\\&=\frac14\left[\lim_{x\to0}\frac{e^x-1}x-\lim_{x\to0}\frac{e^x-1}{-x}\right]^2\tag6\\&=\left[\lim_{x\to0}\frac{e^x-1}x\right]^2\tag7\\&=1^2=1\tag8\end{align}$$
Thus,

$$2L=1\implies L=\frac12$$


Steps:
$(1):$ Simply $(e^x-1-x)+(e^{-x}-1+x)=e^x+e^{-x}-2$
$(2):$ Turns out to be a perfect square.  Who knew?  ;)
$(3):$ We can take the square out.
$(4):$ Let $x\mapsto2x$.
$(5):$ Factor out the $1/2$ (which gets squared) and then split the limit into two, using: $e^x-e^{-x}=(e^x-1)-(e^{-x}-1)$.
$(6):$  Let $x\mapsto-x$ in the second limit.
$(7):$ Combine the resulting limits into one.
$(8):$ Use the well-known and probably given that $\lim_{x\to0}\frac{e^x-1}x=1$.
A: Hint Use the Taylor Series expansion of $e^{x}$, then expand and cancel like-terms.  
