lim x arrow (5 pi)/(6)(cos 2 x-csc 3 x)

To evaluate the limit \(\lim_{x \rightarrow \frac{5\pi}{6}} (\cos 2x - \csc 3x)\), we first need to understand the behavior of each trigonometric function as \(x\) approaches \(\frac{5\pi}{6}\).<br /><br />1. **Evaluate \(\cos 2x\):**<br /><br /> As \(x \rightarrow \frac{5\pi}{6}\), we have:<br /> \[<br /> 2x \rightarrow 2 \times \frac{5\pi}{6} = \frac{5\pi}{3}<br /> \]<br /> The cosine of \(\frac{5\pi}{3}\) is equivalent to the cosine of \(-\frac{\pi}{3}\) (since \(\frac{5\pi}{3} = 2\pi - \frac{\pi}{3}\)), which is:<br /> \[<br /> \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}<br /> \]<br /><br />2. **Evaluate \(\csc 3x\):**<br /><br /> As \(x \rightarrow \frac{5\pi}{6}\), we have:<br /> \[<br /> 3x \rightarrow 3 \times \frac{5\pi}{6} = \frac{15\pi}{6} = \frac{5\pi}{2}<br /> \]<br /> The cosecant function, \(\csc\), is the reciprocal of the sine function. At \(\frac{5\pi}{2}\), the sine function is:<br /> \[<br /> \sin\left(\frac{5\pi}{2}\right) = \sin\left(2\pi + \frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1<br /> \]<br /> Therefore, \(\csc\left(\frac{5\pi}{2}\right) = \frac{1}{\sin\left(\frac{5\pi}{2}\right)} = 1\).<br /><br />Now, substituting these values into the original expression, we get:<br />\[<br />\lim_{x \rightarrow \frac{5\pi}{6}} (\cos 2x - \csc 3x) = \frac{1}{2} - 1 = -\frac{1}{2}<br />\]<br /><br />Thus, the limit is \(-\frac{1}{2}\).