1. **Stating the problem:** We are given a piecewise function: $$f(z) = \begin{cases} z^2 \sin\frac{1}{z} & z \neq 0 \\ 0 & z = 0 \end{cases}$$ We want to understand the behavior of $f(z)$ near $z=0$, especially for $0 < |z| < \delta$, and explain its continuity. 2. **Formula and important rules:** The function involves $z^2$ multiplied by $\sin\frac{1}{z}$. We know: - $\sin\theta$ oscillates between $-1$ and $1$ for all real $\theta$. - Multiplying by $z^2$ (which tends to 0 as $z \to 0$) will affect the amplitude of oscillations. 3. **Intermediate work:** - For $z \neq 0$, $|f(z)| = |z^2 \sin\frac{1}{z}| \leq |z^2| \cdot 1 = z^2$. - As $z \to 0$, $z^2 \to 0$, so $|f(z)| \to 0$. - At $z=0$, $f(0) = 0$ by definition. 4. **Explanation:** - Although $\sin\frac{1}{z}$ oscillates infinitely fast near zero, the factor $z^2$ shrinks the amplitude of these oscillations to zero. - This means the function $f(z)$ approaches 0 as $z$ approaches 0 from either side. - Hence, $f(z)$ is continuous at $z=0$ because: $$\lim_{z \to 0} f(z) = f(0) = 0$$ 5. **Graph interpretation:** - The blue curve $\sin(1/x)$ oscillates rapidly near zero with constant amplitude between $-1$ and $1$. - The red curve $x^2 \sin(1/x)$ oscillates with diminishing amplitude, confined within $y = \pm x^2$, which shrinks to zero near $x=0$. - The condition $0 < |x| < \delta$ indicates we look at values close to zero but not zero itself. **Final conclusion:** The function $f(z)$ is continuous at $z=0$ because the oscillations are damped by $z^2$, making the limit equal to the function value at zero.