How to compute $\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$ without L'Hospital's Rule or series? I came across this problem in an examination for beginners in Calculus: 

Compute $\displaystyle \lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2}$.

It is easy to show that the limit is $1$ by using series or L'Hospital's Rule. But the examination is aimed to students that know only the basics of Calculus (derivatives of power functions and trigonometric functions, Chain Rule, Mean Value Theorem basically). 
How to compute this limit by elementary means?
 A: $$\begin{align} L = &\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2} &\\ = &\lim_{x\to 0}\frac{\sin x^2 \cos x + \sin x \cos x^2-x}{x^2} &\\= &1+ \lim_{x\to 0}\frac{ \sin x \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x + x) \cos x^2-x}{x^2}&\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)\cos x^2 + x(\cos x^2 -1)}{x^2}  &\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)\cos x^2 - x(2\sin^2(x^2/2))}{x^2} \tag{0} &\\= &1+ \lim_{x\to 0}\frac{ (\sin x - x)}{x^2}  \end{align}$$

Now we need to find $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2}$
We know that $\lim_{x\to 0} \dfrac{\sin x - x}{x^3}$ is $\dfrac {-1}{6}$,
Therefore,
$$\lim_{x\to 0}\frac{ (\sin x - x)}{x^2} = \lim_{x \to 0} x \cdot \lim_{x\to 0}\frac{ (\sin x - x)}{x^3} = 0 \cdot \dfrac{-1}{6} = 0 $$ 
Edit :-
An alternative method to prove $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2} = 0$ is given by Paramanand Singh in the comments,
$$\cos x \leq \dfrac{\sin x}{x} \leq 1 \implies  \dfrac{\cos x  - 1}{x}\leq \dfrac{\sin x - x}{x^2} \leq 0$$  
Using seqeeze theorem and the fact that $\lim_{x\to 0}\dfrac{\cos x - 1}{x} = 0$, we get $\lim_{x\to 0}\dfrac{ (\sin x - x)}{x^2} = 0$. 
Therefore $L =1$


$1) :-$ I used the following on step $(0)$, $$\lim_{x\to 0} \dfrac{x\sin^2 (x^2/2)}{x^2} =\lim_{x\to 0} x\left(\dfrac{\sin (x^2/2)}{x}\right)^2 =\lim_{x\to 0} x\left(\dfrac{\sin (x^2/2)}{x^2} x\right)^2 = 0 $$
$2):-$  $\lim_{x\to 0}\dfrac{\sin x - x}{x^3}$ is derived here. (This limit is not required anymore to find $L$)

A: First we prove that $\lim_{x\to 0} \frac{\sin x - x}{x^2} =0$, which can be done similarly as in the case of proving that $\lim_{x\to 0} \frac{\sin x}{x} =1$. We have that:
$$cos(x) \le \frac{\sin x}{x} \le 1 \implies \frac{\cos x - 1}{x} \le \frac{\sin x - x}{x^2} \le 0$$
and so by the Squeeze Theorem the claim is proven, using the well-known $\lim_{x\to 0} \frac{\cos x - 1}{x} =0$
Now using the MVT we have:
$$\lim_{x\to 0}\frac{\sin(x^2+x)-x}{x^2} = \lim_{x \to 0} \frac{\sin x - x}{x^2} +\lim_{x \to 0} \frac{\sin(x^2+x) - \sin(x)}{(x^2 + x) - x}= 0 + \lim_{x \to 0} \cos(c)= \cos(0) = 1$$
As $c \in (x,x^2+x)$ and $x \to 0$
A: Let $u=x^2+x$ and see that
\begin{align}\frac{\sin(x^2+x)-x}{x^2}&=\frac{\sin(x^2+x)-(x^2+x)+x^2}{(x^2+x)^2}(x+1)^2\\&=\frac{\sin(u)-u}{u^2}(x+1)^2+1\\&\to0+1=1\end{align}
where we used
$$\lim_{u\to0}\frac{\sin(u)-u}{u^2}=0$$
(various proofs see here)
