Use the Squeeze Theorem to evaluate \( \lim_{x \to 0} x \cdot \sin\left(\frac{1}{x^2}\right) \).

To evaluate the limit \[ \lim_{x \to 0} x \cdot \sin\left(\frac{1}{x^2}\right), \] we can use the Squeeze Theorem. First, we need to analyze the behavior of the function \( \sin\left(\frac{1}{x^2}\right) \). We know that the sine function is bounded, specifically: \[ -1 \leq \sin\left(\frac{1}{x^2}\right) \leq 1. \] Multiplying the entire inequality by \( x \) (noting that as \( x \to 0 \), \( x \) can be positive or negative, but we will consider the absolute value), we have: \[ -x \leq x \cdot \sin\left(\frac{1}{x^2}\right) \leq x. \] Now, we need to consider the limits of the bounding functions as \( x \to 0 \): 1. The limit of the lower bound \( -x \) as \( x \to 0 \): \[ \lim_{x \to 0} -x = 0. \] 2. The limit of the upper bound \( x \) as \( x \to 0 \): \[ \lim_{x \to 0} x = 0. \] Since both the lower and upper bounds converge to 0, we can apply the Squeeze Theorem. According to the Squeeze Theorem, if \[ f(x) \leq g(x) \leq h(x) \] for all \( x \) in some interval around \( a \) (except possibly at \( a \)), and if \[ \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L, \] then \[ \lim_{x \to a} g(x) = L. \] In our case, we have: \[ -x \leq x \cdot \sin\left(\frac{1}{x^2}\right) \leq x, \] and both limits as \( x \to 0 \) are 0. Therefore, we conclude that: \[ \lim_{x \to 0} x \cdot \sin\left(\frac{1}{x^2}\right) = 0. \] Thus, the final result is: \[ \boxed{0}. \]