Problem 29 - Essential Calculus Early Transcendentals

Question

Find the limit. $$\lim _{x \rightarrow \infty}\left(e^{-2 x} \cos x\right)$$

Step-by-Step Solution

Answer

0

Step 1: Understand the Problem

We are tasked with finding the limit of the function \( e^{-2x} \cos x \) as \( x \) approaches infinity. This involves solving for the behavior of the function when \( x \) becomes very large.

Step 2: Identify Behavior of Each Component

Investigate the behavior of \( e^{-2x} \) and \( \cos x \) separately as \( x \to \infty \). The exponential term \( e^{-2x} \) goes to 0 as \( x \) increases, since the exponent \(-2x\) becomes increasingly negative. Meanwhile, \( \cos x \) oscillates between -1 and 1 and does not settle at a single value.

Step 3: Combine with Product Limit Theorem

The limit of the product \( \,e^{-2x} \cos x \,\) can be assessed by considering the product limit rule: \( \lim_{x \to c} f(x)g(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) \), if both limits exist. Here, \( \lim_{x \to \infty} e^{-2x} = 0 \), while \( \lim_{x \to \infty} \cos x \) does not exist due to oscillation, yet \( e^{-2x} \) dominates by approaching 0.

Step 4: Apply Squeeze Theorem

Use the Squeeze Theorem. Since \( -e^{-2x} \leq e^{-2x} \cos x \leq e^{-2x} \), and both bounds \(-e^{-2x}\) and \(e^{-2x}\) approach 0 as \( x \to \infty \), we can conclude that \( e^{-2x} \cos x \) also tends to 0.

Step 5: Calculate the Limit

Based on the above analysis, \( \,\lim_{x \to \infty} e^{-2x} \cos x = 0 \,\) because the exponential decay \( e^{-2x} \to 0 \) dominates the behavior of the entire expression despite the oscillations in \( \cos x \).

Key Concepts

Exponential Decay

Exponential decay is a situation in which a quantity decreases at a rate proportional to its current value. This is characterized by formulas involving negative exponents, such as \( e^{-2x} \) in our exercise. As \( x \) increases, the exponent - \(-2x\) becomes very large and negative
- This results in the output \( e^{-2x} \) getting closer to zero as \( x \) approaches infinity.
This behavior is crucial in understanding many phenomena in nature, such as radioactive decay and capacitor discharge in electronics.

• The exponential term dictates the overall behavior, overpowering other components in a product, especially as those components might only fluctuate or oscillate without converging to a single value.
• In the given problem, \( e^{-2x} \) approaches zero much faster than \( \cos x \) can alter the result significantly.

Recognizing exponential decay is key when evaluating limits where exponential functions are involved.

Oscillating Functions

Functions like \( \cos x \) demonstrate oscillating behavior. Oscillating functions are those that constantly rise and fall between two values without settling at a single limit.
A clear feature of \( \cos x \) is its movement between -1 and 1 as \( x \) changes. This oscillation means that it does not approach a particular limit as \( x \) goes to infinity. Regardless of \( x \):

• The value of \( \cos x \) will fluctuate and never settle.
• This lack of convergence is common in functions involving trigonometric terms.

In our limit problem, this means - The oscillating nature of \( \cos x \) makes its own limit non-existent over infinity
- But this does not impede the rest of the expression from having a finite limit due to the presence of other dominating factors like \( e^{-2x} \).

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus used to find the limit of a function by "squeezing" it between two functions whose limits are known. It is especially useful when dealing with functions that oscillate or are otherwise complex.
In practice, if you have a function \( f(x) \) that can be bounded by two other functions \( g(x) \) and \( h(x) \) such that \( g(x) \leq f(x) \leq h(x) \) and \( \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L \)
then it follows that \( \lim_{x \to c} f(x) = L \).
For the problem at hand:- We observe that \( e^{-2x} \cos x \) is squeezed between \( -e^{-2x} \) and \( e^{-2x} \).

• As both bounding functions approach zero as \( x \to \infty \), we conclude indeed the limit of our given function \( \lim_{x \to \infty} e^{-2x} \cos x = 0 \).
• The theorem confirms the dominance of exponential decay in this product, even with an oscillating part.

Product of Functions

When evaluating limits of products like \( e^{-2x} \cos x \), it's important to understand how the separate components interact. The product of functions takes into account the limits of individual functions to predict the overall outcome.
Here’s how the interaction works:

• \( e^{-2x} \) goes rapidly to zero.
• \( \cos x \) oscillates, meaning its limit doesn't exist alone.

However, for the product limit to exist:- If one of the functions in multiplication goes to zero and another remains bounded or oscillates
- The product will still approach zero
Using these insights, the exponential decay in our specific task completely guides the behavior of the product, making the whole expression tend towards a limit of zero. This strategy, applying concepts like limits of individual factors, suffices to evaluate complex product limit problems.