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Chapter 1: Problem 5

Find the discontinuities, if any. $$ f(x)=\csc x $$

Short Answer

Expert verified
The function \(f(x) = \csc x\) has discontinuities at \(x = n\pi\), where \(n\) is an integer.

Step by step solution

01

Understand the Function

The function given is the cosecant function, which is defined as the reciprocal of the sine function. So, we have: \[f(x) = \csc x = \frac{1}{\sin x}\] This means the function is not defined wherever \(\sin x = 0\).
02

Find Where \(\sin x = 0\)

The sine function \(\sin x\) equals zero at integer multiples of \(\pi\), i.e., at \(x = n\pi\) where \(n\) is any integer. So at these points, \(\csc x\) is undefined, leading to discontinuities.
03

Classify the Discontinuity

Each discontinuity at \(x = n\pi\) for the function \(\csc x\) is an infinite discontinuity because as \(x\) approaches \(n\pi\), \(f(x) = \csc x\) tends towards \(\pm\infty\). This happens because the sine function approaches zero, making \(\frac{1}{\sin x}\) extremely large in magnitude.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discontinuities
When working with functions, discontinuities are points where a function is not continuous. In simple terms, this means the graph of the function makes a sudden "jump" or "breaks" at those points. For the function \(f(x) = \csc x\), discontinuities occur because the cosecant function is the reciprocal of the sine function. Here鈥檚 why:
  • The cosecant function is defined as \(\csc x = \frac{1}{\sin x}\).
  • Whenever \(\sin x = 0\), the function \(\csc x\) becomes undefined.
  • The sine function equals zero at integer multiples of \(\pi\) (i.e., \(x = n\pi\), where \(n\) is an integer).
At these points, the function is not defined, resulting in discontinuities on the graph of \(f(x) = \csc x\). Understanding discontinuities is crucial for graphing and analyzing trigonometric functions like the cosecant function.
Infinite Discontinuity
Infinite discontinuities are a special kind of discontinuity where the function tends to positive or negative infinity as it approaches certain points. With the cosecant function, we encounter infinite discontinuities at points where \(x = n\pi\), as follows:
  • At these points, \(\sin x \) approaches zero, but never actually reaches it.
  • As \(x\) gets infinitely close to \(n\pi\), the function \(f(x) = \csc x = \frac{1}{\sin x}\) becomes larger in magnitude, heading towards \(+\infty\) or \(-\infty\).
  • On a graph, this is evident by vertical asymptotes or sharp spikes.
These infinite discontinuities are characteristic of the cosecant function's graph, where there is a sharp gap, indicating the function is not defined at those specific points. Grasping this idea helps in predicting and sketching behaviors of various trigonometric functions.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, especially in the study of periodic phenomena. The cosecant function is one of these, defined in terms of the sine function:
  • Cosecant is the reciprocal of sine, expressed as \(\csc x = \frac{1}{\sin x}\).
  • Typical trigonometric functions include sine, cosine, tangent, secant, cosecant, and cotangent.
  • These functions are periodic, which means they repeat their values in regular intervals.
Understanding trigonometric functions is critical, as they form the basis for more complex concepts in calculus, physics, and engineering. They are used in modelling waves, oscillations, and other repetitive patterns seen in nature and technology. The cosecant function, with its distinctive infinite discontinuities and periodic nature, is a perfect example of how intriguing and varied trigonometric functions can be.

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Most popular questions from this chapter

A function \(f\) is said to have a removable discontinuity at \(x=c\) if \(\lim _{x \rightarrow c} f(x)\) exists but \(f\) is not continuous at \(x=c\), either because \(f\) is not defined at \(c\) or because the definition for \(f(c)\) differs from the value of the limit. This terminology will be needed in these exercises. (a) The terminology removable discontinuity is appropriate because a removable discontinuity of a function \(f\) at \(x=c\) can be "removed" by redefining the value of \(f\) appropriately at \(x=c\). What value for \(f(c)\) removes the discontinuity? (b) Show that the following functions have removable discontinuities at \(x=1\), and sketch their graphs. $$ f(x)=\frac{x^{2}-1}{x-1} \quad \text { and } \quad g(x)=\left\\{\begin{array}{ll} 1, & x>1 \\ 0, & x=1 \\ 1, & x<1 \end{array}\right. $$ (c) What values should be assigned to \(f(1)\) and \(g(1)\) to remove the discontinuities?

Find the limits. $$ \lim _{x \rightarrow 3^{+}} \frac{x}{x-3} $$

Find all values of \(a\) such that $$ \lim _{x \rightarrow 1}\left(\frac{1}{x-1}-\frac{a}{x^{2}-1}\right) $$ exists and is finite.

Find the limits. $$ \lim _{y \rightarrow+\infty} \frac{2-y}{\sqrt{7+6 y^{2}}} $$

Two students are discussing the limit of \(\sqrt{x}\) as \(x\) approaches \(0 .\) One student maintains that the limit is 0 , while the other claims that the limit does not exist. Write a short paragraph that discusses the pros and cons of each student's position.
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