Problem 96 - Thomas Calculus in SI Units

Question

Find the limits. $$\lim _{x \rightarrow \infty} \frac{e^{x} \tan ^{-1} e^{x}}{e^{2 x}+x}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Analyze the Function

The function given is \(\lim _{x \rightarrow \infty} \frac{e^{x} \tan ^{-1} e^{x}}{e^{2 x}+x}\). As \(x\) approaches infinity, both the numerator and denominator involve growing exponential functions and polynomial terms. Thus, an asymptotic analysis will help.

Step 2: Asymptotic Behavior of \(\tan^{-1} e^x\)

First, consider \(\tan^{-1} e^x\). As \(x\) approaches infinity, \(e^x\) grows, and \(\tan^{-1} e^x\) approaches \(\frac{\pi}{2}\). This is because \(\tan^{-1}\) of any large positive number approaches \(\frac{\pi}{2}\) as the input approaches infinity.

Step 3: Simplifying the Numerator

Given the asymptotic behavior, the numerator can be approximated by \(e^x \cdot \frac{\pi}{2}\) for large \(x\). This simplifies to \(\frac{\pi}{2} e^x\).

Step 4: Simplifying the Denominator

In the denominator, the expression is \(e^{2x} + x\). As \(x\) approaches infinity, \(e^{2x}\) will dominate \(x\) because exponential functions grow faster than polynomial functions. Thus, the denominator simplifies to approximately \(e^{2x}\).

Step 5: Substituting Asymptotic Approximations into the Limit

Considering the simplified expressions, substitute them into the limit: \(\lim_{x \rightarrow \infty} \frac{\frac{\pi}{2} e^x}{e^{2x}} = \frac{\pi}{2} \cdot \lim_{x \rightarrow \infty} \frac{e^x}{e^{2x}} = \frac{\pi}{2} \cdot \lim_{x \rightarrow \infty} \frac{1}{e^x}\).

Step 6: Evaluate the Final Limit

Since \(\lim_{x \rightarrow \infty} \frac{1}{e^x} = 0\), the entire expression becomes \(\frac{\pi}{2} \cdot 0 = 0\). Therefore, the limit is 0.

Key Concepts

Asymptotic Analysis

Asymptotic analysis is a mathematical technique used to describe the behavior of functions as the input grows very large or very small.
It provides a simplified way to understand complex mathematical functions by focusing on their dominant terms when the variables approach infinity.

• The main idea is to approximate functions in terms of simpler functions that exhibit the same behavior at the extreme values.
• This analysis helps in comparing the growth rates of different functions to find which terms can be ignored as variables tend to infinity.

When analyzing a limit at infinity like in our exercise, asymptotic analysis helps us simplify the problem by identifying the dominant factors.
For example, between polynomial functions and exponential functions, the latter grows much faster and dominates the expression as seen in the denominator of our original problem where we simplified from \(e^{2x} + x\) to \(e^{2x}\).

Exponential Functions

Exponential functions are crucial to calculus due to their unique properties of rapid growth.
An exponential function, commonly expressed as \(e^x\), grows faster than any polynomial function as the value of \(x\) increases.

• This rapid growth plays a significant role in limit calculations, especially at infinity.
• In calculus, exponential functions are often used to represent natural processes like growth and decay.

In the context of our example, the exponential function \(e^x\) is pivotal both in the numerator and the denominator.

• In the numerator, paired with \tan^{-1}e^x\, it represents how fast the input is growing.
• In the denominator, \(e^{2x}\) outpaces the polynomial term \(x\), simplifying our analysis and limit determination.

Limits at Infinity

Limits at infinity deal with the behavior of functions as inputs tend to increasingly large positive or negative values.
They help determine the end behavior of a function and are essential in understanding horizontal asymptotes in graphs.

• In the case of \(x ightarrow \infty\), we observe how the terms of a function behave as \(x\) takes much larger values.
• For limits involving exponential forms, identifying the dominating term is essential to finding the final simplified limit.

For our original problem, \(\lim _{x \rightarrow \infty} \frac{e^{x} \cdot \tan^{-1} e^{x}}{e^{2x} + x}\), analyzing the behavior at infinity tells us that both the numerator and the denominator contain expressions growing indefinitely.
By applying simplifications using limits at infinity, such as observing that \lim_{x \rightarrow \infty} \frac{1}{e^x} = 0\, we determine that the overall limit becomes zero. This illustrates the necessity to observe how each term behaves independently at infinite bounds to make accurate limit evaluations.