Problem 73 - Thomas Calculus in SI Units

Question

L'Hópital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow \infty} \frac{e^{x^{2}}}{x e^{x}}$$

Step-by-Step Solution

Answer

The limit is infinity.

Step 1: Identifying the Limit Form

First, identify the form of the limit as \( x \) approaches infinity: \( \lim_{x \to \infty} \frac{e^{x^2}}{x e^x} \). Evaluate the behavior of the numerator \( e^{x^2} \) and the denominator \( x e^x \) as \( x \to \infty \): Both functions tend to infinity, indicating it is an indeterminate form \( \frac{\infty}{\infty} \).

Step 2: Simplifying the Exponent

Observe that using L'Hôpital's Rule repeatedly doesn't simplify the expression. Instead, rewrite the expression and factor it differently: \[ \lim_{x \to \infty} \frac{e^{x^2}}{x e^x} = \lim_{x \to \infty} \frac{e^{x^2 - x}}{x}. \]

Step 3: Analyzing the New Exponent Expression

Consider the expression in the exponent: \( x^2 - x \). As \( x \to \infty \), the \( x^2 \) term dominates, causing \( x^2 - x \to \infty \). Thus, \( e^{x^2 - x} \to \infty \).

Step 4: Comparison Between Growth Rates

Recognize that \( e^{x^2 - x} \) grows significantly faster than the linear term \( x \) in the denominator. This suggests that the numerator \( e^{x^2 - x} \) approaches infinity much more rapidly than \( x \).

Step 5: Final Limit Evaluation

Given that the numerator \( e^{x^2 - x} \) grows much more rapidly than the linear \( x \) in the denominator, the entire expression \( \frac{e^{x^2 - x}}{x} \) will approach infinity. Thus, the limit is \( \infty \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits that appear in indeterminate forms. It applies when a limit results in forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), where direct substitution doesn't give a clear answer. The rule states that if \(\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0\) or \(\infty\), then:

• \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\)

provided the limit on the right-hand side exists. However, L'Hôpital's Rule is not always effective, especially when applying it repeatedly yields no simplification. In such cases, other algebraic techniques may be required. This is precisely the situation encountered in the given exercise, where L'Hôpital's Rule continues to cycle without resolving the limit. Thus, exploring alternative methods becomes necessary.

Limit Evaluation

Understanding limit evaluation is crucial in calculus, as it helps determine the behavior of functions as they approach specific points or infinity. The goal is to find the value that a function approaches, even if it never actually reaches it. In the exercise, we seek to evaluate:

• \(\lim_{x \to \infty} \frac{e^{x^2}}{x e^x}\)

Initially, it's recognized as an indeterminate form \(\frac{\infty}{\infty}\). Instead of using L'Hôpital's Rule, we can simplify by factoring the expression or analyzing terms' growth rates. Simplification may involve changing how terms are presented, as seen by rewriting \(\frac{e^{x^2}}{x e^x}\) as \(\frac{e^{x^2 - x}}{x}\). This highlights significant terms better and aids in assessing the limit. Ultimately, such evaluations can lead to understanding phenomena in calculus without relying on more complex calculus rules.

Indeterminate Forms

Indeterminate forms are specific cases in calculus where limits result in expressions that do not directly indicate convergence or divergence. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\infty - \infty\), and others. These forms suggest ambiguity in the values functions approach, thereby necessitating different methods for resolution. In the exercise, we encounter the indeterminate form \(\frac{\infty}{\infty}\). This arises because both the numerator \(e^{x^2}\) and the denominator \(x e^x\) increase without bound as \(x\) approaches infinity. To handle such cases, one typically employs techniques like factoring or comparing growth rates. By assessing how fast different parts of the expression grow, we can determine dominance and, thus, evaluate the limit. Here, by rewriting and recognizing the dominance of \(e^{x^2 - x}\) over a linear \(x\), we conclude that the limit moves towards infinity, illustrating the solution strategy for indeterminate forms.