Prove that $\lim _{h \rightarrow 0} \frac{f(2 h)-f(-5 h)-21 h}{h^2} < \infty$ for a $C^2(\Bbb R)$ function $f$ with $f(0)=1,f'(0)=3$, and $f''(0)=5$ Have a look at the multiple choice question:
Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a twice differentiable function such that $f^{\prime \prime}$ is continuous and $f(0)=1, f^{\prime}(0)=$ 3. $f^{\prime \prime}(0)=5$. Which of the following limit always exists ?
(A) $\lim _{h \rightarrow 0} \frac{f(2 h)-f(3 h)+2 h}{h^2}$
(B) $\lim _{h \rightarrow 0} \frac{f(2 h)-f(-5 h)-21 h}{h^2}$
(C) $\lim _{h \rightarrow 0} \frac{f(3 h)+f(-3 h)-2}{h}$
(D) $\lim _{h \rightarrow 0} \frac{f(h)+f(2 h)-2 f(3 h)+2 h}{h^2}$
I have eliminated the options A and D using the function $f(x)=\frac{5}{2}x^2+3x+1$. But I couldn't deal options B and C. How can I use the fact $\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$ here?
 A: Let's prove the option B and C is true by using the definition of derivative and Taylor expansion with piano remainder.
\begin{equation*}
  \begin{aligned}
   &\quad\lim _{h \rightarrow 0} \frac{f(2 h)-f(-5 h)-21 h}{h^2}\\
&=\lim _{h \rightarrow 0} \frac{[1+3\cdot2h+\frac{5}{2}\cdot4h^2+o(4h^2)]-[1-3\cdot5h+\frac{5}{2}\cdot25h^2+o(25h^2)]-21 h}{h^2}\\
   &=\lim _{h \rightarrow 0} \frac{-\frac{5}{2}\cdot21h^2+o(4h^2)+o(25h^2)}{h^2}\\
   &=-\frac{105}{2}.
   \end{aligned}
\end{equation*}
\begin{equation*}
  \begin{aligned}
   \lim _{h \rightarrow 0} \frac{f(3 h)+f(-3 h)-2}{h}&=\lim _{h \rightarrow 0} 3\frac{f(3 h)-f(0)+f(-3 h)-f(0)}{3h}\\
   &=3\lim _{\Delta x \rightarrow 0} \frac{f(\Delta x)-f(0)+f(-\Delta x)-f(0)}{\Delta x}\\
   &=3\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}-3\lim _{\Delta x \rightarrow 0}\frac{f(0-\Delta x)-f(0)}{-\Delta x}\\
   &=0.
   \end{aligned}
\end{equation*}
A: Proving Option (B) is trivial using L'Hospital Rule.
$\displaystyle \frac{f( 2h) -f( -5h) -21h}{h^{2}} \ \ \ ( 0/0\ form) \ \Longrightarrow \ \frac{2f'( 2h) +5f'( -5h) -21}{2h}$
$\displaystyle \frac{2*3+5*3-21}{2*0} \ ( 0/0\ form) \ \Longrightarrow \ \frac{4f''( 2h) \ -25f''( -5h)}{2}$
Substituting the values: we get the limit is $\displaystyle ( 4( 5) -25( 5)) /2\ =\ -52.5\ $
Which is a finite value. Hence option (B) is correct.
Option (C) is correct again by LHospital Rule  as $\displaystyle \frac{3f'( 3h) -3f'( -3h)}{1} \ =\ 0$
