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 \(x\) approaches infinity.<br /><br />First, let's rewrite the expression:<br />\[ x \sin\left(\frac{2}{x}\right). \]<br /><br />As \(x \to \infty\), \(\frac{2}{x} \to 0\), and \(\sin\left(\frac{2}{x}\right) \to \sin(0) = 0\). However, we need to be more precise in our analysis.<br /><br />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 \(x\) in some interval around \(c\), except possibly at \(c\) itself, and if \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\).<br /><br />For \(\sin\left(\frac{2}{x}\right)\), we know that:<br />\[ -1 \leq \sin\left(\frac{2}{x}\right) \leq 1. \]<br /><br />Multiplying this inequality by \(x\), we get:<br />\[ -x \leq x \sin\left(\frac{2}{x}\right) \leq x. \]<br /><br />Now, we can split this into two separate limits:<br />\[ \lim_{x \to \infty} -x = -\infty, \]<br />\[ \lim_{x \to \infty} x = \infty. \]<br /><br />Since \(-x\) and \(x\) sandwich \(x \sin\left(\frac{2}{x}\right)\), and the limits of \(-x\) and \(x\) are \(-\infty\) and \(\infty\) respectively, the Squeeze Theorem does not directly apply here. Instead, we should consider the behavior of the product \(x \sin\left(\frac{2}{x}\right)\).<br /><br />We can use the fact that \(\sin\left(\frac{2}{x}\right) \approx \frac{2}{x}\) for large \(x\):<br />\[ \sin\left(\frac{2}{x}\right) \approx \frac{2}{x}. \]<br /><br />Thus,<br />\[ x \sin\left(\frac{2}{x}\right) \approx x \cdot \frac{2}{x} = 2. \]<br /><br />Therefore, by the Squeeze Theorem and the approximation, we conclude that:<br />\[ \lim_{x \to \infty} x \sin\left(\frac{2}{x}\right) = 2. \]<br /><br />So, the final answer is:<br />\[ \boxed{2}. \]