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Chapter 1: Problem 93

Prove that \(f(x)=\cos x\) is continuous everywhere.

Short Answer

Expert verified
To prove that \(f(x) = \cos x\) is continuous everywhere, we first establish that for a function to be continuous at a point \(x_0\), the limit of the function as \(x\) approaches \(x_0\) must equal the function's value at \(x_0\). To accomplish this, we utilized the difference identity for cosine and rewrote the limit of the difference between cosine functions in terms of sine functions. We then applied the known limit \(\lim_{x\to 0} \frac{\sin x}{x} = 1\) and evaluated the limit to be 0. This shows that the function is continuous at any point, and therefore \(f(x) = \cos x\) is continuous everywhere.

Step by step solution

01

Establish the Definition of the Limit

For a function to be continuous at a point \(x_0\), the limit must exist, and the limit as \(x\) approaches \(x_0\) must be equal to the function's value at \(x_0\). Mathematically, this can be written as: \[ \lim_{x\to x_0} f(x) = f(x_0) \] Our goal is to show that this condition holds for any \(x_0 \in \mathbb{R}\).
02

Use the Difference Identity for Cosine

We can rewrite the limit in terms of the difference between the function values at any given point \(x\) and the point of interest \(x_0\). In order to do this, we will use the difference identity for cosine: \[\cos{A} - \cos{B} = -2\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A - B}{2}\right)\] Let \(A = x\) and \(B = x_0\), then we can write the difference of the function values as: \[\cos x - \cos x_0 = -2\sin \left(\frac{x+x_0}{2}\right) \sin \left(\frac{x-x_0}{2}\right)\]
03

Evaluate the Limit

Now, we rewrite the limit in terms of the difference identity: \[\lim_{x\to x_0} \left(\cos x - \cos x_0\right) = \lim_{x\to x_0} \left[ -2\sin \left(\frac{x+x_0}{2}\right) \sin \left(\frac{x-x_0}{2}\right) \right]\] Notice that the only part of the expression which depends on \(x\) and approaches 0 as \(x \to x_0\) is the term \(\sin \left(\frac{x - x_0}{2}\right)\). We will use the fact that \(\lim_{x\to 0} \frac{\sin x}{x} = 1\), a well-known limit. Now let us make the substitution \(u = x - x_0\), which implies that \(x \to x_0\) when \(u \to 0\). Thus, the limit becomes \[\lim_{u \to 0} \frac{-2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \sin \left(\frac{u}{2}\right)}{u}\] This limit can be separated into two: \[\lim_{u \to 0} \left[ -2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \cdot \lim_{u \to 0} \frac{\sin \left(\frac{u}{2}\right)}{u} \right]\]
04

Apply the Known Limit

Next, apply the known limit \(\lim_{x\to 0} \frac{\sin x}{x} = 1\) to the second limit and simplify: \[\lim_{u \to 0} \left[ -2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \cdot 1 \right] = -2\, \sin \left(\frac{x_0 + x_0}{2}\right) \cdot 0\] The entire expression evaluates to 0: \[\lim_{x\to x_0} (\cos x - \cos x_0) = 0\] This means that the limit of the difference between any two points of the function as they approach each other is 0, which implies that the function is continuous at any point \(x_0\). Therefore, the function \(f(x) = \cos x\) is continuous everywhere.

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Key Concepts

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

Limit of a Function
Understanding the limit of a function is vital in determining its continuity. A limit describes the behavior of a function as the input approaches a certain value. In the case of continuity, for a function to be continuous at a point \(x_0\), the limit of the function as \(x\) approaches \(x_0\) must exist and be equal to the function's value at \(x_0\). This is mathematically represented as \[ \lim_{x\to x_0} f(x) = f(x_0) \].

For the cosine function, \(\cos x\), we examine this limit for every point on the function to ensure it aligns with the value of the function at that point. This ensures that the function is continuous everywhere and does not have any gaps or jumps at any point within its domain.
Difference Identity for Cosine
The difference identity for cosine is an important trigonometric identity that helps in analyzing the behavior of the cosine function. This identity expresses the difference between the cosine of two angles \(A\) and \(B\) as \[ \cos{A} - \cos{B} = -2\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A - B}{2}\right) \].

By applying this identity, we can simplify the expression for the difference between \(\cos x\) and \(\cos x_0\) and thus, work with the function's continuity in a more manageable form. The fact that this identity relates the cosine function with the sine function allows us to then use fundamental limits related to \(\sin x\) in our analysis.
Trigonometric Limits
Trigonometric limits involve understanding how trigonometric functions behave as angles approach a certain value. One of the most fundamental and useful trigonometric limits is \[ \lim_{x\to 0} \frac{\sin x}{x} = 1 \].

This limit is pivotal when dealing with trigonometric functions in calculus. By substituting \(u = x - x_0\), we align our expression with this well-known limit to facilitate our calculations. In the context of the continuity of the cosine function, this trigonometric limit assists in affirming that the limit of the cosine function exists and is equal to the function's value at the point \(x_0\), thus proving the function鈥檚 continuity.
Continuous Functions
A continuous function is one without breaks, jumps, or holes in its graph. In more formal terms, a function is continuous at a point \(x_0\) when the following conditions are met: the function is defined at \(x_0\), the limit of the function as \(x\) approaches \(x_0\) exists, and the value of the limit equals the function鈥檚 value at \(x_0\). If these conditions are met for every point in a function's domain, then the function is continuous everywhere.

In the context of the cosine function, we've demonstrated through the limit and difference identity that at any given point \(x_0\), the function maintains continuity since all criteria are met. Thus, \(\cos x\) is a model example of a continuous function.

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Most popular questions from this chapter

Let \(g\) be a continuous function on an interval \([a, b]\) and suppose \(a \leq g(x) \leq b\) whenever \(a \leq x \leq b .\) Show that the equation \(x=g(x)\) has at least one solution \(c\) in the interval \([a, b] .\) Give a geometric interpretation. Hint: Apply the Intermediate Value Theorem to the function \(f(x)=x-g(x)\)

The position function of an object moving along a straight line is given by \(s=f(t) .\) The average velocity of the object over the time interval \([a, b]\) is the average rate of change of f over \([a, b] ;\) its (instantaneous) velocity at \(t=a\) is the rate of change of \(\bar{f}\) at \(a .\) Velocity of a Car Suppose the distance \(s\) (in feet) covered by a car moving along a straight road after \(t\) sec is given by the function \(s=f(t)=2 t^{2}+48 t\). a. Calculate the average velocity of the car over the time intervals \([20,21],[20,20.1]\), and \([20,20.01]\). b. Calculate the (instantaneous) velocity of the car when \(t=20 .\) c. Compare the results of part (a) with those of part (b).

Prove the Constant Multiple Law for limits: If \(\lim _{x \rightarrow a} f(x)=L\) and \(c\) is a constant, then \(\lim _{x \rightarrow \alpha}[c f(x)]=c L .\)

The symbol [ ] denotes the greatest integer function defined by \([x]=\) the greatest integer \(n\) such that \(n \leq x .\) For example, \([2.8]=2\), and \([-2.7]=-3 .\) In Exercises \(23-28\), use the graph of the function to find the indicated limit, if it exists. \(\lim _{x \rightarrow-1}[x]\)

Average Velocity of a Helicopter A helicopter lifts vertically from its pad and reaches a height of \(h(t)=0.2 t^{3}\) feet after \(t\) sec, where \(0 \leq t \leq 12\). a. How long does it take for the helicopter to reach an altitude of \(200 \mathrm{ft}\) ? b. What is the average velocity of the helicopter during the time it takes to attain this height? c. What is the velocity of the helicopter when it reaches this height?
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