Determine the interval(s) on which the following function is continuous, then analyze the given limits. \[ h(x)=\frac{3+3 \sin x}{\cos x} ; \lim _{x \rightarrow x / 2^{-}} h(x) ; \lim _{x \rightarrow 4 x / 3} h(x) \] A. \( \lim _{x \rightarrow x / 2^{-}} \frac{3+3 \sin x}{\cos x}=\infty \) (Type an exact answer, using radicals as needed.) B. The limit does not exist and is neither infinity nor negative infinity. Select the correct choice below, if necessary, fill in the answer box to complete your choice. A. \( \lim _{x \rightarrow 4 x / 3} \frac{3+3 \sin x}{\cos x}= \) \( \square \) (Type an exact answer, using radicals as needed.) B. The limit does not exist and is neither infinity nor negative infinity.

The function \( h(x) = \frac{3 + 3 \sin x}{\cos x} \) is continuous wherever the denominator is not equal to zero. Therefore, \( h(x) \) is continuous on the intervals \( x \neq (2n + 1)\frac{\pi}{2} \), where \( n \) is any integer, which corresponds to the points where \( \cos x = 0 \). Now, let’s analyze the limits! As \( x \) approaches \( \frac{\pi}{2} \) from the left, the denominator \( \cos x \) approaches 0, leading \( h(x) \) to approach infinity. Thus, \( \lim_{x \rightarrow \frac{\pi}{2}^{-}} h(x) = \infty \). Moving on to the limit as \( x \) approaches \( \frac{4\pi}{3} \): since \( \cos \left(\frac{4\pi}{3}\right) \) is negative and \( 3 + 3 \sin\left(\frac{4\pi}{3}\right) \) computes to a non-zero value, the limit exists. Plugging in the values gives \( 3 + 3(-\frac{\sqrt{3}}{2}) = 3 - \frac{3\sqrt{3}}{2} \). The limit can be computed as \[ \lim_{x \rightarrow \frac{4\pi}{3}} h(x) = \frac{3 - \frac{3\sqrt{3}}{2}}{-\frac{1}{2}} = -6 + 3\sqrt{3} \] So, choose option A, and fill in the answer box with \( -6 + 3\sqrt{3} \).