4. HaǎTu mpertest: lim _(xarrow infty )(sin3x)/(sin5x)

To evaluate the limit \(\lim_{x \to \infty} \frac{\sin(3x)}{\sin(5x)}\), we can use the fact that \(\sin(kx)\) for any constant \(k\) oscillates between -1 and 1 as \(x\) increases. Specifically, \(\sin(kx)\) does not settle down to a single value but continues to oscillate.<br /><br />To see this more formally, consider the behavior of the sine functions:<br /><br />\[<br />\sin(3x) \text{ and } \sin(5x)<br />\]<br /><br />Both \(\sin(3x)\) and \(\sin(5x)\) oscillate between -1 and 1 as \(x\) increases. Since the sine function oscillates indefinitely, the ratio \(\frac{\sin(3x)}{\sin(5x)}\) does not converge to a single finite value. Instead, it continues to oscillate.<br /><br />To make this more precise, let's consider the period of the sine functions. The period of \(\sin(3x)\) is \(\frac{2\pi}{3}\), and the period of \(\sin(5x)\) is \(\frac{2\pi}{5}\). As \(x\) approaches infinity, both \(\sin(3x)\) and \(\sin(5x)\) will continue to oscillate without settling down.<br /><br />Given this, we can conclude that:<br /><br />\[<br />\lim_{x \to \infty} \frac{\sin(3x)}{\sin(5x)}<br />\]<br /><br />does not exist because the ratio continues to oscillate indefinitely. Therefore, the limit is undefined.<br /><br />So, the final answer is:<br /><br />\[<br />\lim_{x \to \infty} \frac{\sin(3x)}{\sin(5x)} \text{ does not exist.}<br />\]