Question

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

Short answer

Question: Evaluate the given limits for the function $$f(x) = \csc x$$: 1. $$\lim _{x \rightarrow \pi / 4} f(x)$$ 2. $$\lim _{x \rightarrow 2 \pi^{-}} f(x)$$ Answer: 1. $$\lim _{x \rightarrow \pi / 4} f(x) = \sqrt{2}$$ 2. $$\lim _{x \rightarrow 2 \pi^{-}} f(x) = +\infty$$

Step-by-step solution

Identify the properties of the given function

The function $$ f(x) = \csc x $$ is the reciprocal of the sine function ($$ \csc x = \frac{1}{\sin x} $$). Since the sine function is continuous for all real numbers, we must analyze the values for which $$\sin x \neq 0$$ (Because we can't divide by zero).

Determine the intervals for continuous function

The sine function $$\sin x$$ has zero values at $$x = n\pi$$, where n is an integer. So, the cosecant function $$f(x) = \csc x$$ will be undefined at these points. To find the continuous intervals, we need to exclude these points from the domain of real numbers: The continuous intervals for $$f(x) = \csc x$$ are $$\displaystyle\bigcup_{n=-\infty}^{\infty} (n\pi, (n+1)\pi)$$.

Evaluate the limits

Now, let us evaluate the given limits. Limit 1: $$\lim _{x \rightarrow \pi / 4} f(x)$$ Since the function $$f(x) = \csc x$$ is continuous on the interval $$\left(\frac{\pi}{2}, \pi \right)$$, which contains $$\pi/4$$, we can directly substitute it: $$\lim _{x \rightarrow \pi / 4} f(x) = \csc (\pi / 4) = \frac{1}{\sin(\pi / 4)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$$ Limit 2: $$\lim _{x \rightarrow 2 \pi^{-}} f(x)$$ For this limit, $$x$$ is approaching $$2\pi$$ from the left. As $$x$$ gets close to $$2\pi$$, $$\sin x$$ approaches zero, and $$f(x) = \csc x = \frac{1}{\sin x}$$ becomes very large (positive Infinity). So, $$\lim _{x \rightarrow 2 \pi^{-}} f(x) = +\infty$$. In conclusion, the function $$f(x) = \csc x$$ is continuous on the interval $$\displaystyle\bigcup_{n=-\infty}^{\infty} (n\pi, (n+1)\pi)$$, and the given limits are: $$\lim _{x \rightarrow \pi / 4} f(x) = \sqrt{2}$$ $$\lim _{x \rightarrow 2 \pi^{-}} f(x) = +\infty$$

Key concepts

Cosecant Function

The cosecant function, denoted as \( \text{csc}(x) \) or \( \frac{1}{\text{sin}(x)} \), is a trigonometric function that is defined as the reciprocal of the sine function. Trigonometric functions like cosecant play a prominent role in various fields such as physics, engineering, and mathematics. Understanding the behavior of the cosecant function is crucial for solving many problems in trigonometry.

Specifically, \( \text{csc}(x) \) is undefined when \( \text{sin}(x) = 0 \) because division by zero is not permissible in mathematics. The sine function is zero at multiples of \( \text{\(\pi\)} \), which means the cosecant function will be undefined at these points. Thus, when examining the continuity and limits of the cosecant function, it is essential to consider these undefined points and consider intervals where \( \text{sin}(x) \) does not equal zero.

Limits of Functions

The concept of limits is a fundamental aspect of calculus and analysis which deals with the behavior of functions as they approach a particular point. A limit attempts to describe the value that a function approaches as the input approaches some value. For the limit \( \text{lim}_{x \to c} f(x) \), the goal is to find the value that \( f(x) \) gets arbitrarily close to as \( x \) approaches \( c \).

For continuous functions, evaluating the limit at a point where the function is defined is often straightforward—we simply substitute the value into the function. However, when dealing with functions that are not continuous, such as the cosecant function near the points where sine is zero, limits can describe the infinite behavior (like \( +\text{\(\infty\)} \)) or oscillating behavior (for functions that don’t settle to a single value).

Intervals of Continuity

An interval of continuity refers to a range of input values over which a function is continuous. For a function to be continuous at a point, it must fulfill three conditions: the function is defined at the point, the limit exists at that point, and the limit at that point equals the function’s value. To determine intervals of continuity for a function like cosecant, we identify where the function is not continuous and exclude those points from the domain.

In the case of \( f(x) = \text{csc}(x) \), we consider the intervals between the multiples of \( \text{\(\pi\)} \) where \( \text{sin}(x) \) is not zero. These intervals are in the form of \( (n\text{\(\pi\)}, (n+1)\text{\(\pi\)}) \) for all integers \( n \), giving us a clear picture of where the function maintains continuity. Students should pay close attention to the end behavior as the function approaches points of discontinuity, which is a common exam topic.