Problem 17 - Calculus Single Variable

Question

In each of Exercises \(17-24,\) apply l'Hôpital's Rule repeatedly (when needed) to evaluate the given limit, if it exists. \(\lim _{x \rightarrow \infty} x^{2} / e^{3 x}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form of the Limit

First, determine the form of the limit as \(x \to \infty\). The limit is \( \lim_{x \to \infty} \frac{x^2}{e^{3x}} \). As \(x\) approaches infinity, \(x^2\) goes to infinity and \(e^{3x}\) also goes to infinity. This produces an indeterminate form of \(\frac{\infty}{\infty}\), which means l'Hôpital's Rule can be applied.

Step 2: Apply l'Hôpital's Rule (First Time)

Since the limit is in the indeterminate form \(\frac{\infty}{\infty}\), apply l'Hôpital's Rule. Take derivatives of the numerator and the denominator:Derivative of the numerator \(x^2\) is \(2x\).Derivative of \(e^{3x}\) is \(3e^{3x}\).Thus, apply l'Hôpital's Rule:\[\lim_{x \to \infty} \frac{x^2}{e^{3x}} = \lim_{x \to \infty} \frac{2x}{3e^{3x}}\]

Step 3: Apply l'Hôpital's Rule (Second Time)

Again, analyze the limit \(\lim_{x \to \infty} \frac{2x}{3e^{3x}}\). This is still in the form \(\frac{\infty}{\infty}\). Apply l'Hôpital's Rule once more. Take derivatives:Derivative of \(2x\) is \(2\).Derivative of \(3e^{3x}\) is \(9e^{3x}\).Thus, apply l'Hôpital's Rule:\[\lim_{x \to \infty} \frac{2x}{3e^{3x}} = \lim_{x \to \infty} \frac{2}{9e^{3x}}\]

Step 4: Evaluate the Final Limit

Evaluate the expression \(\lim_{x \to \infty} \frac{2}{9e^{3x}}\). As \(x\) approaches infinity, \(e^{3x}\) grows very rapidly towards infinity, thus \(9e^{3x}\) also approaches infinity. Therefore, the fraction \(\frac{2}{9e^{3x}}\) approaches \(0\).

Key Concepts

Indeterminate Form

When evaluating limits in calculus, we often encounter expressions that initially seem challenging to resolve, especially those that don't fit neatly into a defined outcome. These are known as indeterminate forms. The expression itself doesn't plainly indicate whether the limit exists. There are several types of indeterminate forms, such as:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \times \infty\)
• \(\infty - \infty\)

In our specific exercise, \(\lim_{x \rightarrow \infty} x^{2} / e^{3x}\) gives the indeterminate form \(\frac{\infty}{\infty}\) when evaluated at \(x \to \infty\). This form often arises, prompting us to use techniques like l'Hôpital's Rule to find the limit.

Limit Evaluation

Finding the limit of a function as \(x\) approaches a particular value involves comprehending the behavior of the function nearing that point. In essence, it tells us about the tendency of function values. When evaluating limits where direct substitution leads to an indeterminate form like \(\frac{\infty}{\infty}\), we can't directly ascertain the limit.
In such cases, l'Hôpital's Rule becomes an invaluable tool. The rule allows us to replace the original limit with the limit of the derivatives of the numerator and denominator, given that they also form an indeterminate form. It's crucial to re-evaluate the limit after each application of the rule, as it may necessitate multiple applications if the form persists.

Derivatives Application

In calculus, derivatives provide us with the rate at which a function changes. When dealing with limits exhibiting indeterminate forms, l'Hôpital's Rule utilizes derivatives to simplify the evaluation. By deriving both the numerator and the denominator, we transform complex terms into manageable ones.
For example, with \(\lim_{x \to \infty} \frac{x^2}{e^{3x}}\), applying l'Hôpital's Rule requires us to differentiate \(x^2\) and \(e^{3x}\). The derivative of \(x^2\) is \(2x\), and that of \(e^{3x}\) is \(3e^{3x}\). Repeated derivations may be needed, as seen when we derive again obtaining \(2\) and \(9e^{3x}\), respectively.

• The result aids in simplifying the limit evaluation.
• Each derivative calms the chaos of infinity in the expressions.

Infinite Limits

Infinite limits occur when a function's output grows without bound as \(x\) approaches a particular value. When \(x\) trends towards infinity, understanding the behavior of expressions like exponential functions or polynomials is key. These functions expand vastly as their inputs increase.In the exercise \[\lim_{x \rightarrow \infty} \frac{2}{9e^{3x}}\], \(e^{3x}\) climbs rapidly toward infinity, forcing the denominator to grow considerably larger than the constant numerator. This results in the fraction approaching \(0\), as a smaller numerator over an infinite denominator always trends towards zero.

• This showcases how exponential terms typically overpower polynomial terms in limits at infinity.
• Recognizing this behavior aids in quickly evaluating such expressions without excessive computation.