b) lim _(xarrow infty )xsin((2)/(x))=

To evaluate the limit \(\lim_{x \to \infty} x \sin\left(\frac{2}{x}\right)\), we can use the fact that \(\sin\left(\frac{2}{x}\right)\) approaches 0 as approaches infinity.First, let's rewrite the expression: As , , and \(\sin\left(\frac{2}{x}\right) \to \sin(0) = 0\). However, we need to be more precise in our analysis.We can use the Squeeze Theorem to evaluate this limit. The Squeeze Theorem states that if \(f(x) \leq g(x) \leq h(x)\) for all in some interval around , except possibly at itself, and if \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\).For \(\sin\left(\frac{2}{x}\right)\), we know that: Multiplying this inequality by , we get: Now, we can split this into two separate limits: Since and sandwich \(x \sin\left(\frac{2}{x}\right)\), and the limits of and are and respectively, the Squeeze Theorem does not directly apply here. Instead, we should consider the behavior of the product \(x \sin\left(\frac{2}{x}\right)\).We can use the fact that \(\sin\left(\frac{2}{x}\right) \approx \frac{2}{x}\) for large : Thus, Therefore, by the Squeeze Theorem and the approximation, we conclude that: So, the final answer is: