How can we compute $\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3}$ and $\lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$? Could we compute the limits 
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$ 
without using the l'Hospital rule and the Taylor expansion?
 A: Hint. If one recalls that, for any function $f$ differentiable  near $0$,
$$
\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0) \tag1
$$ then one may write
$$
\begin{align}
l:=\lim_{x \to 0}\frac{e^x-\sin (x)-1}{x^2}=\lim_{x \to 0}\:\frac1x\left(\frac{e^x-1}{x} -\frac{\sin x}x\right)
\end{align}
$$ then one may apply $(1)$ to
$$
f(x)=\frac{e^x-1}{x} -\frac{\sin x}x,\quad f(0)=0,
$$$$
f'(x)=-\frac{e^x-\sin (x)-1}{x^2}+\frac{e^x}{x} -\frac{\cos x}x,
$$ getting
$$
l=-l+\lim_{x \to  0}\frac{e^x-1}{x}-\lim_{x \to  0}\frac{\cos x-1}{x}
$$ and
$$
l=-l+1-0
$$ that is

$$
l=\lim_{x \to  0}\frac{e^x-\sin (x)-1}{x^2}=\frac12.
$$

Can you take it from here applying it to the first limit?
A: The first one is equal to $1+\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}.$ We can apply to this limit the technique in this answer of mine. 
Let $L=\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}$. This is not infinite if it exists because near $0$ one has $$-\frac12\leftarrow\frac{\cos x-1}{x^2}<\frac{\frac{\sin x}{x}-1}{x^2}<\frac{1-1}{x^2}=0.$$ Then, considering a similar limit, $$\lim_{x\to0}\frac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\frac{\sin(2x)-2x}{2x^3}=4L$$ we can deduce$$\lim_{x\to0}\frac{\sin x\cos x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12=3L,$$that is $L=-\frac16.$ So your limit equals $\frac56$.
Getting back to this; due to the finiteness of the first limit, the second one is the same as $$\lim_{x\to0}\frac{e^x-1-x}{x^2}+\lim_{x\to0}\frac{x-\sin x}{x^2}=\lim_{x\to0}\frac{\frac{e^x-1}{x}-1}{x},$$i.e. $f'(0)$ where $f(x)=\frac{e^x-1}{x}, f(0)=1.$ One has $$f'(0)=\lim_{x\to0}\frac{e^x(x-1)+1}{x^2}=\lim_{x\to0}\frac{e^x-1}{x}-\lim_{x\to0}\frac{e^x-1-x}{x^2}=1-f'(0),$$ so it equals $\frac12$.
Alternatively, let $L_2$ be this limit. It is largely known that $e^x>1+x$; this also means $e^{-x}>1-x$, or equivalently for $x<1$, $e^x<\frac1{1-x},$ and it's easy to prove $\frac1{1-x}<1+x+2x^2 $ for $0\ne x<\frac12.$ Thus, $0\le L_2\le2.$ Assuming it exists, we have $$L_2=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{2x}-(1+x)^2}{x^2(e^x+1+x)}=\frac12\lim_{x\to0} \frac{e^{2x}-1-2x-x^2}{x^2}=2L_2-\frac12,$$whence $L_2=\frac12$.
A: Sure: use Wolfram Alpha or Maple or ...
