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!
$$\lim\limits_{x\rightarrow 0^+}{\dfrac{e^x-\sin x-1}{x^2}}$$
 A: 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:
$$\frac{1}{2}-x\leq\frac{e^x-\sin x-1}{x^2}\leq \frac{1}{2}+x$$
$$\lim_{x\to 0^+}\frac{e^x-\sin x-1}{x^2}=\frac{1}{2}$$
A: $$\lim_{x \to 0^+}\frac{e^x-\sin x -1}{x^2}=\lim_{x \to 0^+}\frac{e^x- x -1}{x^2}+\lim_{x \to 0^+}\frac{x-\sin x}{x^2}$$
First part given here
 and next part $$|\frac{x -\sin x}{x^2}-0|  < \frac{x}{x^2}< \epsilon  \implies x >\frac{1}{\epsilon} $$ Take $N(\epsilon)= \lfloor  \frac{1}{\epsilon}\rfloor+1$ So here from definition of limit we get $$\lim_{x \to 0^+}\frac{x-\sin x}{x^2} =0$$
 Otherwise see this
A: I have an idea that might be fruitful although I haven't been very careful and completely ignore the $x\rightarrow 0^+$! Define the complex function 
$\displaystyle F(z)=\frac{e^z-\sin z-1}{z^2}$.
Now let $z=it$.
$\displaystyle  F(z)=\frac{e^{it}-\sin(ti)-1}{t^2}=\frac{\cos t+i\sin t-\sin(ti)-1}{t^2}$
$\displaystyle \Rightarrow  F(z)=\frac{\cos t-1}{t^2}+\frac{i\sin t-\sin(it)}{t^2}$.
Now take the limit as $t\rightarrow0$. It is not difficult to show that the first limit is one half (multiply above and below by $\cos t+1$). I do not know how to compute the second limit which is annoying me although I think that it is zero.
Now use the fact that 
$\displaystyle \lim_{x\rightarrow 0}\,F(x)=\lim_{z\rightarrow 0}\,F(z)=\lim_{t\rightarrow 0}\,F(it)$
I have assumed that the middle limit exists which is troublesome.
A: Perhaps the problem says $x\to 0^+$ because some inequalities wich relate $\sin x$ and $e^x$ with polynomials are only valid for $x>0$ sufficiently close to $0$. So, a strategy is to find functions $f(x)$ and $g(x)$ such that
$$f(x)\leq \dfrac{e^x-\sin x-1}{x^2}\leq g(x)\:\;\mbox{ for }x>0 \mbox{ close to } 0$$
and $\displaystyle\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}g(x)=\frac{1}{2}$
