Find limit without using L'hopital or Taylor's series

I'm trying to solve this limit $without$ using L'hopital's Rule or Taylor Series. Any help is appreciated!

$$\lim\limits_{x\rightarrow 0^+}{\dfrac{e^x-\sin x-1}{x^2}}$$

Answer

One possible way is to shoot linear functions at the limit - not very elegant, but it works. Let:

$$f(x)=x^3-\frac{x^2}{2}+e^x-\sin x-1,\;\;x\geq 0$$

Computing the first few derivatives of $f:$

$$f'(x)=3x^2-x+e^x-\cos x$$
$$f''(x)=6x-1+e^x+\sin x$$

$f''$ is clearly increasing and since $f''(0)=0$ we have $f''(x)>0$ for $x\in (0,a)$ for some $a$.
This in turn implies that $f'$ is strictly increasing and since $f'(0)=0$ we again have $f'(x)>0$ for $x\in (0,a)$. Finally, this means $f$ is also increasing on this interval, and since $f(0)=0$ we have:

$$0\leq x\leq a:\quad f(x)\geq 0$$

$$\Rightarrow \;\;\frac{e^x-\sin x-1}{x^2}\geq \frac{1}{2}-x$$

Similarly by considering $h(x)=-x^3-\dfrac{x^2}{2}+e^x-\sin x-1$ it is very easy to show that:

$$0\leq x\leq b: \quad h(x)\leq 0$$

$$\Rightarrow \;\;\frac{1}{2}+x\geq\frac{e^x-\sin x-1}{x^2}$$

Hence for small positive $x$ we have:

$$\frac{1}{2}-x\leq\frac{e^x-\sin x-1}{x^2}\leq \frac{1}{2}+x$$

$$\lim_{x\to 0^+}\frac{e^x-\sin x-1}{x^2}=\frac{1}{2}$$