Evaluate $\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$ without L'Hôpital This limit is one of the "Problems Plus" from Stewart Calculus:
$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$
Note that the limit is of the indeterminate form $\frac{0}{0}$. The problem appears several chapters before L'Hôpital's rule is discussed, so I assume there is a solution without using L'Hopital.
Looking at a graph, the local behavior of the function near $0$ appears to be $-\sin(a+x)$, which, of course, suggests a limit of $-\sin(a)$. Using L'Hôpital's rule twice confirms this guess:
$$\begin{align}
& \lim_{x \to 0} \frac{\frac{d}{dx} [\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)]}{ \frac{d}{dx} x^{2}} \\
& = \lim_{x \to 0} \frac{2\cos(a+2x)-2\cos(a+x)}{2x} \\
& = \lim_{x \to 0} \frac{-4\sin(a+2x)+2\sin(a+x)}{2} \\
& = -\sin(a).
\end{align}$$
Can anyone give a hint or solution for evaluating this limit without L'Hôpital?

Without heavy machinery, only that $\lim_{x \to 0} \sin x / x = 1$, you can use the sum-to-product formulas, as follows.
\begin{align*}
\sin(a+2x) - 2 \sin(a+x) + \sin a & = \sin(a+2x) - \sin(a+x) + \sin a - \sin(a+x) \\
& = 2 \sin(x/2) \cos(a + 3x/2) - 2 \sin (x/2) \cos(a + x/2) \\
& = 2 \sin(x/2) \left (\cos(a + 3x/2) - \cos(a + x/2) \right ) \\
& = - 4 \sin(x/2) \sin(a+x) \sin(x/2) \\
& = - 4 \sin^2(x/2) \sin(a+x).
\end{align*}
Then
$$
\lim_{x \to 0} \frac{\sin^2(x/2)}{x^2} = \frac14 \lim_{x \to 0} \frac{\sin^2(x/2)}{(x/2)^2} = \frac14 \lim_{y \to 0} \left ( \frac{\sin y}{y} \right )^2 = \frac14,
$$
and finally the limit of the expression you are looking at is indeed
$$
-4 \times \frac14 \times \sin a = - \sin a.
$$
