In order to evaluate the limit \( \lim _{x \rightarrow 0}|x| \sin \left(\frac{1}{x^{2}-x}\right) \) we are required to use the Squeeze theorem. When creating bounds for the limit, which of the following functions represents an appropriate upper bound that allows us to use the Squeeze theorem.

To evaluate the limit using the Squeeze theorem, we can consider that \( |\sin(t)| \leq 1 \) for all real numbers \( t \). Therefore, we can bound the sine term in your limit. This gives us: \[ |x|\sin\left(\frac{1}{x^{2}-x}\right) \leq |x| \cdot 1 = |x| \] Thus, an appropriate upper bound for the limit is \( |x| \). To apply the Squeeze theorem effectively, we also need a lower bound. Since \( \sin\left(\frac{1}{x^{2}-x}\right) \) oscillates between -1 and 1, we can conclude that: \[ -|x| \leq |x|\sin\left(\frac{1}{x^{2}-x}\right) \leq |x| \] As \( x \) approaches 0, both the upper and lower bounds tend to 0, allowing us to conclude that the limit itself must also tend to 0. Using the Squeeze theorem, we find that: \[ \lim_{x \to 0} |x|\sin\left(\frac{1}{x^{2}-x}\right) = 0. \] Isn’t that a neat application of squeezing? It’s like putting a big ol’ hug on the function right at the crucial moment!