Determine the interval(s) on which the following functions are continuous; then analyze the given limits.

f(x)=1+sin x / cos x; limx→π/2^− f(x); lim x→4π/3 f(x)

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The function given is \( f(x) = 1 + \frac{\sin x}{\cos x} \). This can be rewritten as \( f(x) = 1 + \tan x \) since \( \tan x = \frac{\sin x}{\cos x} \).

The function \( \tan x \) is continuous wherever \( \cos x \neq 0 \). The cosine function is zero at odd multiples of \( \frac{\pi}{2} \), i.e., \( x = \frac{\pi}{2} + k\pi \) where \( k \) is an integer.

Since \( f(x) = 1 + \tan x \), it is continuous on intervals where \( \tan x \) is continuous. Therefore, \( f(x) \) is continuous on intervals of the form \( (k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2}) \) for integer \( k \).

As \( x \) approaches \( \pi/2 \) from the left, \( \tan x \to +\infty \) because \( \cos x \to 0^+ \). Therefore, \( f(x) = 1 + \tan x \to +\infty \).

At \( x = 4\pi/3 \), \( \cos(4\pi/3) \neq 0 \), so \( \tan(4\pi/3) \) is defined. Calculate \( \tan(4\pi/3) \) and substitute into \( f(x) = 1 + \tan x \) to find the limit.>

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