Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 0} x \cos (1 / x) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the problem

We need to find the limit of the function \( f(x) = x \cos\left(\frac{1}{x}\right) \) as \( x \to 0 \). The goal is to determine if the function approaches a specific value or if it behaves erratically.

Step 2: Analyze the behavior of \( \cos\left(\frac{1}{x}\right) \)

The function \( \cos\left(\frac{1}{x}\right) \) oscillates between \(-1\) and \(1\) as \( x \) approaches 0 because \( \frac{1}{x} \) goes to infinity. This causes the cosine function to keep oscillating without settling on a particular value.

Step 3: Estimate the limit using the Squeeze Theorem

To apply the Squeeze Theorem, note that \( -1 \leq \cos\left(\frac{1}{x}\right) \leq 1 \). Therefore, \(-|x| \leq x\cos\left(\frac{1}{x}\right) \leq |x|\). As \( x \rightarrow 0 \), both \(-|x|\) and \(|x|\) approach 0. This squeezes \( x\cos\left(\frac{1}{x}\right) \) to 0.

Step 4: Conclude the result

Since \( x\cos\left(\frac{1}{x}\right) \) is squeezed between \(-|x|\) and \(|x|\), and both boundaries go to 0 as \( x \rightarrow 0 \), by the Squeeze Theorem, we conclude that\( \lim _{x \rightarrow 0} x \cos (1 / x) = 0 \).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus used to evaluate limits of functions. Imagine being stuck between two narrower and narrower roads that are both heading towards the same destination. This is what the Squeeze Theorem essentially helps us understand when it comes to limits. If you have a function that is difficult to tackle directly, but it's squeezed between two simpler functions whose limits are known, then the limit of the more complicated function will be the same as the limits of the simpler ones.

In our problem, we use this theorem to evaluate the limit of the function \(x \cos\left(\frac{1}{x}\right)\) as \(x\) approaches zero. We know that \(-1 \leq \cos\left(\frac{1}{x}\right) \leq 1\), which allows us to write:

• \(-|x| \leq x \cos\left(\frac{1}{x}\right) \leq |x|\).

Here, both \(-|x|\) and \(|x|\) approach zero as \(x \rightarrow 0\). Thus, according to the Squeeze Theorem, \(x \cos\left(\frac{1}{x}\right)\) is squeezed to the value of 0.

Trigonometric Limits

Trigonometric limits often involve understanding the behavior of trigonometric functions as their arguments approach certain points. These functions can be tricky, especially when dealing with infinity or oscillations. Let's focus on the cosine function, a repeating wave form that oscillates between -1 and 1.

In the exercise, the function inside the cosine, \(\frac{1}{x}\), approaches either positive or negative infinity as \(x\) approaches 0. The limit of a trigonometric function like cosine when its argument is going to infinity isn't straightforward on its own. This is because, without constraints, \(\cos(1/x)\) never settles on one particular value due to its periodic nature.

• Key limit fact: If the argument inside a trigonometric function goes to infinity, the limit of the function can become undefined without additional context.

Understanding this behavior helps us estimate and evaluate limits in trigonometric expressions that have arguments approaching infinity.

Oscillating Functions

Oscillating functions, such as the cosine or sine functions, swing between bounds without settling on a specific value. These functions pose a unique challenge when calculating limits, as they do not easily converge to a single number. Instead, they hover back and forth, making direct evaluation complicated.

For instance, in the function \(x \cos\left(\frac{1}{x}\right)\), \(\cos\left(\frac{1}{x}\right)\) oscillates rapidly as \(x\) approaches zero because \(\frac{1}{x}\) moves toward infinity. However, despite the oscillation of the cosine term, the multiplicative effect of \(x\) helps contain this behavior.

• When we multiply \(\cos(1/x)\) by \(x\), the influence of \(x\) causes \(x \cos\left(\frac{1}{x}\right)\) to be bounded and eventually tend to zero.

Understanding how oscillating and bounding behaviors work together is crucial. These insights help demystify why the function remains bounded and demonstrates why the limit exists despite the irregular behavior of \(\cos\left(\frac{1}{x}\right)\).