Expansion of function , defined on a open interval containing $0$ , in terms of $\sin$ function Let $I$ be an open interval in $\mathbb R$ containing $0$ and $f:I \to \mathbb R$ be a twice differentiable function , then is it true that $$\lim_{x \to  0}\dfrac {f(x)-f(0)-f'(0)\sin x - \dfrac{f''(0)\sin^2 x}2 }{x^2}=0 \space \space...(i)$$ ? I can prove the result if it is additionally given that $f''$ is continuous at $0$ , because then , for some $\theta \in (0,1) $ , we can write by Maclauraine's formula $\dfrac {f(x)-f(0)-f'(0)\sin x - \dfrac{f''(0)\sin^2 x}2 }{x^2}=\dfrac {(x-\sin x)f'(0) +\dfrac{f''(\theta x)}2x^2-\dfrac{f''(0)}2\sin^2x }{x^2}$ , now 
by L'Hospital's rule  it can easily be checked that $\lim_{x\to 0}\dfrac{x-\sin x}{x^2}=0$ and we also know that $\lim_{x\to 0}\dfrac{\sin^2x}{x^2}=1$ and if $f''$ is continuous at $0$ then we could write $\lim_{x\to 0}f''(\theta x)=f''(0)$ and we would be done but if it is not given that $f''(0)$ is continuous at $0$ then can we still infer about the limit $(i)$ ?
 A: $$
f(x)-f(0)-f'(0)\sin x-\frac{f''(0)\sin^2 x}2=\\
f(x)-f(0)-f'(0)x-\frac{f''(0)x^2}2+f'(0)(x-\sin x)+\frac{f''(0)}2(x^2-\sin^2x)
$$
and
$$
\lim_{x\to0}\frac{1}{x^2}\Bigl(f(x)-f(0)-f'(0)x-\frac{f''(0)x^2}2\Bigr)=0,
$$
$$
\lim_{x\to0}\frac{1}{x^2}(f'(0)(x-\sin x))=0,
$$
$$
\lim_{x\to0}\frac{1}{x^2}\Bigl(\frac{f''(0)}2(x^2-\sin^2x)\Bigr)=0.
$$
The first equality follows from the Peano form of the remainder in Taylor's theorem.
A: Let's try L'Hospital's Rule $$\begin{aligned}L &= \lim_{x \to 0}\dfrac{f(x) - f(0) - f'(0)\sin x - \dfrac{f''(0)}{2}\sin^{2}x}{x^{2}}\\
&= \lim_{x \to 0}\dfrac{f'(x) - f'(0)\cos x - f''(0)\sin x\cos x}{2x}\text{ (by L'Hospital's Rule)}\\
&= \frac{1}{2}\lim_{x \to 0}\left(\frac{f'(x) - f'(0)}{x} + f'(0)\cdot\frac{1 - \cos x}{x} - f''(0)\frac{\sin x}{x}\cdot \cos x\right)\\
&= \frac{1}{2}\left(f''(0) + f'(0)\cdot 0 - f''(0)\cdot 1\cdot 1\right) = 0\end{aligned}$$ Note that we have not used continuity of $f''(x)$ just its existence at $x = 0$. Also we have used $$\lim_{x \to 0}\frac{1 - \cos x}{x} = \lim_{x \to 0}\frac{1 - \cos^{2} x}{x(1 + \cos x)} = \lim_{x \to 0}\sin x\cdot\frac{\sin x}{x}\cdot\frac{1}{1 + \cos x} = 0\cdot 1\cdot\frac{1}{2} = 0$$
