We know that: \[ -1 \leq \sin\left(\frac{1}{x}\right) \leq 1, \] which implies: \[ e^{-1} \leq e^{\sin\left(\frac{1}{x}\right)} \leq e^1. \] Multiplying through by \( x^2 \), where \( x^2 \geq 0 \): \[ x^2 e^{-1} \leq x^2 e^{\sin\left(\frac{1}{x}\right)} \leq x^2 e^1. \] Taking the limit as \( x \to 0 \): \[ \lim_{x \to 0} x^2 e^{-1} = 0, \quad \lim_{x \to 0} x^2 e^1 = 0. \] By the Squeeze Theorem: \[ \lim_{x \to 0} x^2 e^{\sin\left(\frac{1}{x}\right)} = 0. \] Thus: \[ \lim_{x \to 0} x^2 e^{\sin\left(\frac{1}{x}\right)} = 0. \]