L'H with functional equation

Question

I am working on learning limits, and I decided to create a derivative of a function, convert it to a limit, and try to solve it as a limit.

Here's my equation: $$\lim_{d \to 0} \frac{2\sin(x+d)-\sin(x)-\sin(x+2d)}{d^2} = f(x)$$

By the way, I know $f(x) = \sin(x)$.

My first idea is to do L'Hopital's Rule, but I don't know how to do that with a limit on $d$ going to $0$ and a $x$ doing... nothing. I have tried multiple things, both of which screw up the limit to not be $\sin(x)$. Does anyone know the correct way to apply L'H to this?

Answer 1

Use L'Hopitals rule, and since $x$ is doing nothing... treat it as a constant, while differentiating w.r. t. $d$.

\begin{align} {\rm L} &=\lim_{d \to 0} \frac{2\sin(x+d)-\sin(x)-\sin(x+2d)}{d^2} \\ &=\lim_{d \to 0} \frac{2\cos(x+d)-2\cos(x+2d)}{2d}\\ &=\lim_{d \to 0} \frac{-\sin(x+d)+2\sin(x+2d)}{1} \\ &= 2\sin x-\sin x\\ &=\color{blue}{\sin x}\\ \end{align}

Answer 2

Another solution using l'Hospital only once:

$$\begin{aligned} & \lim_{d \to 0} \frac{2\sin(x+d)-\sin(x)-\sin(x+2d)}{d^2} \\ &= \lim_{d \to 0} \frac{\cos(x+d)-\cos(x+2d)}{d} \\ &= \lim_{d \to 0} \frac{-2\sin\left(\frac{(x+d)+(x+2d)}{2}\right)\sin\left(\frac{(x+d)-(x+2d)}{2}\right)}{d} \\ &= \lim_{d \to 0} \frac{-2\sin\left(x+\frac{d}{2}\right)\sin\left(-\frac{d}{2}\right)}{d} \\ &= \lim_{d \to 0} \sin\left(x+\frac{d}{2}\right) \cdot \frac{\sin\left(\frac{d}{2}\right)}{\frac{d}{2}} \\ &= \sin(x) \end{aligned}$$

Answer 3

Use Taylor expansion at $x$: $$ \sin(x+d)=\sin x+\frac{\sin'x}{1!}d+\frac{\sin''x}{2!}d^2+o(d^2)= \sin x+d\cos x-\frac{d^2}{2}\sin x+o(d^2) $$ Hence $$ \sin(x+2d)=\sin x+2d\cos x-2d^2\sin x+o(d^2) $$ and finally \begin{align} &2\sin(x+d)-\sin x-\sin(x+2d)\\ &\quad=2\sin x+2d\cos x-d^2\sin x-\sin x-\sin x-2d\cos x+2d^2\sin x+o(d^2)\\ &\quad=d^2\sin x+o(d^2) \end{align} yielding $$ \lim_{d\to0}\frac{2\sin(x+d)-\sin x-\sin(x+2d)}{d^2}= \lim_{d\to0}\frac{d^2\sin x+o(d^2)}{d^2}=\sin x $$