Evaluate Odd Function Value from Definite Integral Ratio Limit

To find the value of $f\left(\frac{1}{2}\right)$, we analyze the properties of the continuous function $f(x)$ and the defined integral functions $F(x)$ and $G(x)$.

Step 1: Properties of $f(x)$

1. $f : \mathbb{R} \to \mathbb{R}$ is a continuous odd function, which implies: $f(-x) = -f(x) \quad \text{for all } x \in \mathbb{R}$
2. Since $f$ is odd, $f(0) = -f(0) \implies f(0) = 0$.
3. We are given that $f(x)$ vanishes at exactly one point. Therefore, $x = 0$ is the only root of $f(x) = 0$, so $f(x) \neq 0$ for all $x \neq 0$.
4. We are given $f(1) = \frac{1}{2} > 0$. By the Intermediate Value Theorem, since $f$ is continuous and does not cross zero on $(0, \infty)$, it must be strictly positive for all $x > 0$: $f(x) > 0 \quad \text{for all } x > 0$ Similarly, $f(x) < 0$ for all $x < 0$.

Step 2: Evaluating $F(1)$ and $G(1)$

The given functions are: $F(x) = \int_{-1}^{x} f(t) \, dt$ $G(x) = \int_{-1}^{x} t |f(f(t))| \, dt$

At $x = 1$:

1. $F(1) = \int_{-1}^{1} f(t) \, dt$. Since $f(t)$ is an odd function, the integral over the symmetric interval $[-1, 1]$ is zero: $F(1) = 0$

2. $G(1) = \int_{-1}^{1} t |f(f(t))| \, dt$. Let $h(t) = t |f(f(t))|$. Evaluating $h(-t)$: $h(-t) = (-t) |f(f(-t))| = -t |f(-f(t))| = -t |-f(f(t))| = -t |f(f(t))| = -h(t)$ Thus, $h(t)$ is also an odd function, so its integral over $[-1, 1]$ is zero: $G(1) = 0$

Step 3: Applying L'Hôpital's Rule

Since $\lim_{x \to 1} F(x) = F(1) = 0$ and $\lim_{x \to 1} G(x) = G(1) = 0$, the limit $\lim_{x \to 1} \frac{F(x)}{G(x)}$ is of the indeterminate form $\frac{0}{0}$.

By the Fundamental Theorem of Calculus: $F'(x) = f(x)$ $G'(x) = x |f(f(x))|$

Applying L'Hôpital's Rule: $\lim_{x \to 1} \frac{F(x)}{G(x)} = \lim_{x \to 1} \frac{F'(x)}{G'(x)} = \lim_{x \to 1} \frac{f(x)}{x |f(f(x))|}$

Step 4: Evaluating the Limit

Since $f(x)$ is continuous at $x = 1$: $\lim_{x \to 1} f(x) = f(1) = \frac{1}{2}$ $\lim_{x \to 1} x |f(f(x))| = 1 \cdot |f(f(1))| = \left|f\left(\frac{1}{2}\right)\right|$

Thus: $\lim_{x \to 1} \frac{F(x)}{G(x)} = \frac{\frac{1}{2}}{\left|f\left(\frac{1}{2}\right)\right|}$

We are given that this limit equals $\frac{1}{14}$: $\frac{\frac{1}{2}}{\left|f\left(\frac{1}{2}\right)\right|} = \frac{1}{14} \implies \left|f\left(\frac{1}{2}\right)\right| = 7$

Since $\frac{1}{2} > 0$, by the positivity property established in Step 1, $f\left(\frac{1}{2}\right) > 0$. Hence: $f\left(\frac{1}{2}\right) = 7$