Evaluate $\displaystyle\lim_{x \to 0} {\sin(a+2x)-2\sin(a+x)+\sin(a)\over x^2}$ I am struggling with this one, I know there is probably a super easy way to solve it but its just slipping my mind. I am trying to stay away from L'Hospital's and the Mean Value Theorem because we studied them later. My teacher refuses to do any examples in class and I am having a hard time keeping up with what he is doing since he is just stating calculus facts and giving calculus history lessons instead of actually teaching. If you could explain that would be super helpful for my future studies! 
 A: Hint: To solve this question without Hospital, use the second derivative
set $a+x=u$ and use Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$
A: You can do this by using the result
$$\lim_{x\to0}\frac{\sin x}x=1$$
without needing to use L'Hopital or MVT, or to write the expression as a derivative.  Using a trig identity for $\sin(a+2x)+\sin a$, your expression is
$$\eqalign{
  \frac{2\sin(a+x)\cos x-2\sin(a+x)}{x^2}
  &=2\sin(a+x)\frac{\cos x-1}{x^2}\cr
  &=-2\sin(a+x)\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}\cr
  &=-2\frac{\sin(a+x)}{1+\cos x}\Bigl(\frac{\sin x}x\Bigr)^2\cr}$$
which tends to $-\sin a$ as $x\to0$.
A: \begin{align*}
\lim_{x\to 0} {\sin(a+2x)-2\sin(a+x)+\sin(a)\over x^2}&= \lim_{x\to 0} {(\sin(a+2x)-\sin(a+x))-(\sin(a+x)-\sin(a))\over x^2}\\
&=\lim_{x\to 0} {2\cos((a+3x/2)\sin(x/2)-(2\cos(a+x/2)\sin(x/2)\over x^2}\\
&= \lim_{x\to 0} {2\sin(x/2)(\cos((a+3x/2)-\cos(a+x/2))\over x^2}\\
&=\lim_{x\to 0} {2\sin(x/2)(2\sin(a+x)\sin(-x/2))\over x^2}\\
&=\lim_{x\to 0} 2\frac{{\sin(x/2)}}{2(x/2)}\frac{\sin(-x/2)}{2(x/2)}2\sin(a)\\
&=-\sin(a)
\end{align*}
A: Let's write $L = \frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2}$. We have
\begin{eqnarray*}
L &=& \frac{\sin(a)\cos(2x)+\sin(2x)\cos(a)-2(\sin(a)\cos(x)+\sin(x)\cos(a))+\sin(a)}{x^2} \\
&=& \frac{\sin(a)(\cos^2(x)-\sin^2(x))+2\sin(x)\cos(a)-2\sin(a)\cos(x)-2\sin(x)\cos(a)+\sin(a)}{x^2} \\
&=& \frac{\sin(a)(1-2\sin^2(x)-2\cos(x)+1)+\cos(a)(2\sin(x)\cos(x)-2\sin(x))}{x^2}\\
&=& 2\sin(a)\bigg[\Big(\frac{1-\cos(x)}{x^2}\Big)-\frac{\sin^2(x)}{x^2}\bigg]+2\cos(a)\bigg[\frac{\sin(x)}{x}\frac{\cos(x)-1}{x}\bigg]
\end{eqnarray*}
So, using the fundamental limit $\displaystyle{\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1}$, one can easily show, by multiplying both numerator and denominator by $1+\cos(x)$, that $\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos(x)}{x} = 0}$ and that $\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}}$.
So, finally, taking limits we obtain that $$\lim_{x \rightarrow 0} \frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2} = 2\sin(a)\Big[\frac{1}{2}-1\Big]+2\cos(a)\cdot 0 \cdot 1 = -\sin(a).$$
A: Noting that
$$
\begin{align}
\lim_{x\to0}\frac{\cos(x)-1}{x^2}
&=\lim_{x\to0}\frac{\cos(x)-1}{\sin^2(x)}\left(\frac{\sin(x)}x{}\right)^2\\
&=\lim_{x\to0}\frac{\cos(x)-1}{1-\cos^2(x)}\ \lim_{x\to0}\left(\frac{\sin(x)}x{}\right)^2\\
&=\lim_{x\to0}\frac{-1}{1+\cos(x)}\ \lim_{x\to0}\left(\frac{\sin(x)}x{}\right)^2\\
&=-\frac12\cdot1^2\tag{1}
\end{align}
$$
and
$$
\begin{align}
&\frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2}\\
&=\sin(a)\frac{\cos(2x)-2\cos(x)+1}{x^2}
+\cos(a)\frac{\sin(2x)-2\sin(x)}{x^2}\\
&=\sin(a)\frac{2\cos^2(x)-2\cos(x)}{x^2}
+\cos(a)\frac{2\sin(x)\cos(x)-2\sin(x)}{x^2}\\
&=2\sin(a)\cos(x)\frac{\cos(x)-1}{x^2}+2\cos(a)\sin(x)\frac{\cos(x)-1}{x^2}\tag{2}
\end{align}
$$
we can take the limit as $x\to0$ to get
$$
\lim_{x\to0}\frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2}=-\sin(a)\tag{3}
$$
