Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous odd function, which vanishes exactly at one point and $f(1) = \frac{1}{2}$. Suppose that $F(x) = \int_{-1}^{x} f(t)\,dt$ for all $x \in [-1, 2]$ and $G(x) = \int_{-1}^{x} t\,|f(f(t))|\,dt$ for all $x \in [-1, 2]$. If $\lim_{x \to 1} \frac{F(x)}{G(x)} = \frac{1}{14}$, then the value of $f\left(\frac{1}{2}\right)$ is

$f \text{ continuous, odd}, \ f(x)=0 \text{ only at } x=0, \ f(1)=\tfrac{1}{2}$ $F(x)=\int_{-1}^{x} f(t)\,dt, \quad G(x)=\int_{-1}^{x} t\,\big|f(f(t))\big|\,dt$ $\lim_{x\to 1}\dfrac{F(x)}{G(x)}=\dfrac{1}{14} \ \Rightarrow \ f\!\left(\tfrac{1}{2}\right)=?$ $1.\ F(1),\ G(1) = 0? \Rightarrow \tfrac{0}{0} \text{ form}$ $2.\ \text{use L'Hôpital}$ $3.\ \dfrac{F'(1)}{G'(1)}=\dfrac{1}{14} \Rightarrow f\!\left(\tfrac{1}{2}\right)$ $F(1)=\int_{-1}^{1} f(t)\,dt$ $f \text{ odd, symmetric interval} \Rightarrow F(1)=0$ $\phi(t)=t\,\big|f(f(t))\big|$ $\phi(-t)=-t\,|f(-f(t))|=-t\,|f(f(t))|=-\phi(t) \Rightarrow G(1)=0$ $\lim_{x\to 1}\dfrac{F(x)}{G(x)} \overset{\text{L'H}}{=} \lim_{x\to 1}\dfrac{F'(x)}{G'(x)}$ $F'(x)=f(x), \quad G'(x)=x\,\big|f(f(x))\big|$ $\dfrac{F'(1)}{G'(1)}=\dfrac{f(1)}{1\cdot|f(f(1))|}=\dfrac{1/2}{\,|f(1/2)|\,}$ $\dfrac{1/2}{|f(1/2)|}=\dfrac{1}{14} \Rightarrow |f(1/2)|=\dfrac{1/2}{1/14}=7$ $f(1/2)=\pm 7$ $f \text{ odd, only zero at } 0, \ f(1)>0 \Rightarrow f(x)>0 \ (x>0)$ $\therefore f(1/2)=7$ $\boxed{f\!\left(\tfrac{1}{2}\right)=7}$