1. **Stating the problem:** We want to analyze the function $$h(x) = \arctan(\sin(\frac{1}{x^2}))$$ and understand its behavior, especially near the y-axis where $x$ approaches 0. 2. **Formula and important rules:** The function is a composition of three parts: - The inner function $f(x) = \frac{1}{x^2}$ which grows very large as $x$ approaches 0. - The sine function $\sin(\theta)$ which oscillates between -1 and 1. - The arctangent function $\arctan(y)$ which maps any real number $y$ to an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. 3. **Intermediate work and explanation:** - As $x \to 0$, $\frac{1}{x^2} \to +\infty$, so $\sin(\frac{1}{x^2})$ oscillates increasingly rapidly between -1 and 1. - Since $\arctan$ is continuous and bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, the output $h(x)$ oscillates between $\arctan(-1) \approx -0.785$ and $\arctan(1) \approx 0.785$. - For large $|x|$, $\frac{1}{x^2}$ is small, so $\sin(\frac{1}{x^2}) \approx \frac{1}{x^2}$ and $h(x) \approx \arctan(\frac{1}{x^2})$, which approaches 0 as $x \to \pm \infty$. 4. **Summary:** The graph oscillates rapidly near $x=0$ between approximately $-0.785$ and $0.785$, and flattens out near 0 for large $|x|$. **Final answer:** The function $h(x) = \arctan(\sin(\frac{1}{x^2}))$ oscillates increasingly rapidly near zero with values bounded between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$, and approaches 0 as $x$ goes to infinity or negative infinity.