5) Use the Squeeze Theorem to show that \( \lim _{x \rightarrow 0} x \sin \frac{1}{x}=0 \)

To show that \( \lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0 \) using the Squeeze Theorem, we start with the fact that the sine function is bounded between -1 and 1. This gives us the inequality: \[ -1 \leq \sin \frac{1}{x} \leq 1 \] Multiplying all parts of this inequality by \( x \) (taking care that \( x \) is approaching 0, where we can consider both positive and negative cases), we get: \[ -x \leq x \sin \frac{1}{x} \leq x \quad \text{for } x > 0 \] \[ x \leq x \sin \frac{1}{x} \leq -x \quad \text{for } x < 0 \] Now, we can rewrite this as two inequalities for \( x \) approaching 0: \[ -x \leq x \sin \frac{1}{x} \leq x \] Next, as \( x \) approaches 0, both bounds \( -x \) and \( x \) approach 0. Using the Squeeze Theorem, since \( x \sin \frac{1}{x} \) is squeezed between two expressions that both converge to 0, we conclude: \[ \lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0 \] Thus, we have used the Squeeze Theorem to establish our result!