Problem 73 - University Calculus: Early Transcendentals

Question

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 infinite.

Step 1: Identify the form of the limit

First, observe the form of the limit. The limit \(\lim _{x \rightarrow \infty} \frac{e^{x^{2}}}{x e^{x}}\) is of the form \(\frac{\infty}{\infty}\). This suggests that L'Hôpital's Rule could be applied after taking derivatives of the numerator and denominator.

Step 2: Apply L'Hôpital's Rule

Since the limit is of type \(\frac{\infty}{\infty}\), apply L'Hôpital's Rule, which states that \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided this limit exists. Calculate the derivatives: - Numerator: \(e^{x^2}\)'s derivative is \(2x e^{x^2}\). - Denominator: \(x e^{x}\)'s derivative is \(e^{x} + x e^{x}\).

Step 3: Re-evaluate the limit

The limit now becomes: \[\lim _{x \rightarrow \infty} \frac{2x e^{x^{2}}}{e^x + xe^x}\]Notice again, as \(x \to \infty\), the expression is still indeterminate of the form \(\frac{\infty}{\infty}\). Apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule again

Take derivatives once more:- For the numerator \(2x e^{x^2}\), use the product rule: first derivative is \((2 + 4x^2) e^{x^2}\).- For the denominator \(e^x + xe^x\), derivative is \((1 + x) e^x + x e^x = (1 + 2x) e^x\).

Step 5: Simplify and evaluate final limit

The limit now becomes: \[\lim _{x \rightarrow \infty} \frac{(2+4x^2)e^{x^2}}{(1+2x)e^x}\]As \(x \to \infty\), notice that the exponential term in the numerator \(e^{x^2}\) grows much faster than \(e^x\) in the denominator. Thus, the fraction grows indefinitely.

Key Concepts

Limit of a Function

The concept of the limit of a function is pivotal in calculus, providing insights into the behavior of functions as the input approaches a specific value. Here, we're interested in what happens to a function as the variable gets infinitely large, \(x \to \infty\). Calculating the limit helps us understand trends and ultimate outcomes.

• The task is to determine the behavior of \( \lim_{x \to \infty} \frac{e^{x^{2}}}{x e^{x}} \).
• We'll use limits to assess how the entire function behaves, especially when direct evaluation isn't straightforward.

In this type of problem, the direct substitution method fails because both the numerator \(e^{x^2}\) and denominator \(xe^x\) approach infinity, creating an indeterminate form \(\frac{\infty}{\infty}\).
This situation calls for a more advanced technique like L'Hôpital's Rule.

Indeterminate Forms

Indeterminate forms occur when the limit computation leads to a fraction that could have multiple potential values, typically appearing in forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
In our exercise, we encounter the indeterminate form \(\frac{\infty}{\infty}\) as both parts of our fraction head towards infinity when \(x \to \infty\).

• Since such forms do not definitively speak to the limit's value, we need further analysis.
• Indeterminate forms signal the need for applying theorems or rules such as L'Hôpital's Rule to resolve the true behavior of the limit.

Here, resolving the indeterminate form through L'Hôpital's Rule involves differentiating the numerator and denominator.
This method effectively transforms the original limit problem into a potentially simpler one, allowing us to find the limit of the derivative expressions.

Derivative Calculations

Derivative calculations are a critical tool in applying L'Hôpital's Rule, which simplifies limits of indeterminate forms by dealing directly with rates of change.
To solve this exercise, we first differentiate the numerator \(e^{x^{2}}\) to get \(2x e^{x^{2}}\) and the denominator \(x e^{x}\) giving us \(e^{x} + x e^{x}\).

• This transforms the indeterminate limit into: \(\lim_{x \to \infty} \frac{2x e^{x^{2}}}{e^{x} + xe^{x}}\).
• After re-evaluating, we find it remains an indeterminate form, prompting another application of L'Hôpital's Rule.
• On repeating the differentiation, the numerator becomes \((2 + 4x^2) e^{x^2}\) and the denominator \((1 + 2x) e^x\).

Further simplification reveals the exponential growth of the numerator outpaces the denominator, reinforcing the conclusion that the limit diverges to infinity.
Mastering derivative calculations ensures efficacy in applying tools like L'Hôpital's Rule.