Question

Determine the interval(s) on which the following functions are continuous; then analyze the given limits. $$f(x)=\frac{1+\sin x}{\cos x} ; \lim _{x \rightarrow \pi / 2^{-}} f(x) ; \lim _{x \rightarrow 4 \pi / 3} f(x)$$

Short answer

Question: Determine the intervals of continuity and analyze the limit of the function f(x) = (1 + sin(x))/cos(x) as x approaches π/2 from the negative side and 4π/3. Answer: The function f(x) is continuous on the intervals $$(-\infty, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{2}) \cup ...$$ The limit of the function as x approaches π/2 from the negative side is ∞, and the limit of the function as x approaches 4π/3 is √3 - 2.

Step-by-step solution

Determine the intervals of continuity

First, we check for points where the denominator is not equal to zero, since f(x) is continuous for all points except where the denominator, cos(x), is zero. For cos(x) = = 0, we get, x = π/2 + nπ , where n is an integer. Therefore, the function f(x) is continuous on the intervals: $$(-\infty, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{2}) \cup ...$$

Calculate the limit as x approaches π/2 from the negative side

We need to consider the limit of the function as x approaches π/2 from the left: $$\lim_{x \rightarrow \pi / 2^{-}} f(x) = \lim_{x \rightarrow \pi / 2^{-}} \frac{1+\sin x}{\cos x}$$ As x approaches π/2 from the left, sin(x) approaches 1, and cos(x) approaches zero (0) from a positive direction. Therefore, the limit becomes: $$\lim_{x \rightarrow \pi / 2^{-}} \frac{1+\sin x}{\cos x} = \frac{1+1}{0^{+}} = \infty$$

Calculate the limit as x approaches 4π/3

Now we consider the limit of the function as x approaches 4π/3: $$\lim_{x \rightarrow 4 \pi / 3} f(x) = \lim_{x \rightarrow 4 \pi / 3} \frac{1+\sin x}{\cos x}$$ As x approaches 4π/3, sin(x) = -√3/2, and cos(x) = -1/2. Therefore, the limit becomes: $$\lim_{x \rightarrow 4 \pi / 3} \frac{1+\sin x}{\cos x} = \frac{1+(-\sqrt{3}/2)}{-1/2} = -\frac{2}{1}+\sqrt{3} = \sqrt{3} - 2$$

Key concepts

Limits of Functions

Limits are fundamental in understanding the behavior of functions as they approach specific points. A limit describes the value a function approaches as the input approaches a certain point. In the context of this problem, we need to calculate the limits of the function \( f(x) = \frac{1+\sin x}{\cos x} \) at certain values of \( x \).
For \( \lim_{x \rightarrow \pi/2^-} f(x) \), we examine how the function behaves as \( x \) approaches \( \pi/2 \) from the left. Here, \( \sin(x) \rightarrow 1 \) and \( \cos(x) \rightarrow 0^+ \), leading to the limit \( \infty \) as the denominator approaches zero from the positive side, implying an infinite limit.
Similarly, for \( \lim_{x \rightarrow 4\pi/3} f(x) \), we substitute the values for sine and cosine at \( 4\pi/3 \), which are \( \sin(4\pi/3) = -\sqrt{3}/2 \) and \( \cos(4\pi/3) = -1/2 \). Evaluating this, the function tends toward a finite number \( \sqrt{3} - 2 \).
Understanding these behaviors helps in analyzing the overall nature and the continuity of the function.

Trigonometric Functions

Trigonometric functions such as \( \sin(x) \) and \( \cos(x) \) are periodic functions that describe relationships within right-angled triangles and the unit circle. These functions oscillate between -1 and 1, and are continuous everywhere, except for individual values where division by zero might occur, like when \( \cos(x) = 0 \).
In our specific function \( f(x) = \frac{1+\sin x}{\cos x} \), the sine component \( \sin x \) applies a vertical shift and adding 1 offsets it. This influences the range of the numerator from 0 to 2.
The denominator, \( \cos x \), determines where the function might have discontinuities. Every time \( \cos x = 0 \), the function is undefined at those points. Thus, understanding trigonometric behavior is key to identifying intervals of continuity as it indicates potential points of discontinuity.

Intervals of Continuity

Determining the intervals on which a function is continuous involves finding where it is not defined or where it might have discontinuities. For rational functions like \( f(x) = \frac{1+\sin x}{\cos x} \), a primary source of discontinuity is values making the denominator zero.
To find such intervals, set \( \cos x = 0 \). This equation is satisfied at points \( x = \pi/2 + n\pi \), where \( n \) is an integer. Consequently, these are the points where \( f(x) \) is discontinuous.
Thus \( f(x) \) is continuous in intervals between these points, which can be expressed as \( (-\infty, \pi/2) \cup (\pi/2, 3\pi/2) \cup (3\pi/2, 5\pi/2) \cup ... \). Recognizing these interval sets helps in fully understanding the continuity of \( f(x) \) excluding specific discontinuous points.