Problem 93 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

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 1: 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 \(x_0 \in \mathbb{R}\).

Step 2: 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: \[\cos x - \cos x_0 = -2\sin \left(\frac{x+x_0}{2}\right) \sin \left(\frac{x-x_0}{2}\right)\]

Step 3: 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: \[\lim_{u \to 0} \left[ -2\, \sin \left(\frac{x_0 + x_0 + u}{2}\right) \cdot \lim_{u \to 0} \frac{\sin \left(\frac{u}{2}\right)}{u} \right]\]

Step 4: 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.

Key Concepts

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’s 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’s 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.