Problem 18 - Calculus Single Variable

Question

Apply l'Hôpital's Rule repeatedly (when needed) to evaluate the given limit, if it exists. \(\lim _{x \rightarrow \infty} x^{3} / e^{2 x}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Verify the Indeterminate Form

To apply l'Hôpital's Rule, we first need to check if the limit is in an indeterminate form such as \( \frac{\infty}{\infty} \). As \( x \to \infty \), both \( x^3 \) and \( e^{2x} \) tend to infinity, so we have the indeterminate form \( \frac{\infty}{\infty} \).

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

According to l'Hôpital's Rule, we differentiate the numerator and the denominator separately. The derivative of \( x^3 \) is \(3x^2\), and the derivative of \( e^{2x} \) is \(2e^{2x}\). Then the limit becomes: \[ \lim _{x \rightarrow \infty} \frac{3x^2}{2e^{2x}} \] This still forms \( \frac{\infty}{\infty} \); hence, l'Hôpital's Rule applies again.

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

Differentiate \(3x^2\) to get \(6x\), and differentiate \(2e^{2x}\) to get \(4e^{2x}\). The limit is now: \[ \lim _{x \rightarrow \infty} \frac{6x}{4e^{2x}} \] This also gives us \( \frac{\infty}{\infty} \). One more application of l'Hôpital's Rule is needed.

Step 4: Apply l'Hôpital's Rule Third Time

Differentiate \(6x\) to get \(6\), and differentiate \(4e^{2x}\) to get \(8e^{2x}\). The limit becomes: \[ \lim _{x \rightarrow \infty} \frac{6}{8e^{2x}} \] Now, as \( x \to \infty \), \( e^{2x} \) tends to infinity, making the whole expression tend towards zero.

Step 5: Evaluate the Final Limit

Since the expression now is \( \frac{6}{8e^{2x}} \) and \( e^{2x} \to \infty \) as \( x \to \infty \), the limit evaluates to 0.

Key Concepts

Indeterminate Forms

Indeterminate forms arise in calculus where the limits result in expressions that are not easily defined. They suggest an undefined state, often requiring special techniques like L'Hôpital's Rule to resolve. A common indeterminate form is \( \frac{\infty}{\infty} \), where both the numerator and denominator approach infinity. This form appears in various limit problems, necessitating further analysis to determine the actual value.

In the original exercise, as \( x \to \infty \), both the numerator \( x^3 \) and the denominator \( e^{2x} \) grow without bound. This results in the indeterminate form \( \frac{\infty}{\infty} \). To handle this situation, L'Hôpital's Rule is applied multiple times to differentiate and simplify the limit, shifting from the \( \frac{\infty}{\infty} \) form to something more determinable. Thus, recognizing and handling indeterminate forms is a crucial skill for successfully evaluating limits.

Limits at Infinity

When we talk about limits at infinity, we refer to what happens to a function's value as the variable approaches infinity. This concept helps us understand the behavior of functions over their entire domain, particularly when the output becomes exceedingly large.

In the context of the given exercise, the limit \( \lim _{x \rightarrow \infty} \frac{x^{3}}{e^{2x}} \) implies observing the behavior of the function as \( x \) grows indefinitely. The polynomial \( x^3 \) grows large, but the exponential \( e^{2x} \) grows even faster, which hints that \( \frac{x^3}{e^{2x}} \) might shrink to zero. By applying L'Hôpital's Rule, we confirm this understanding, eventually showing that the polynomial is outpaced by the exponential function, resulting in the limit being 0.

Differentiation Techniques

Differentiation techniques play an essential role when using L'Hôpital's Rule, where functions are differentiated to resolve indeterminate forms like \( \frac{\infty}{\infty} \). When a limit involves such a form, L'Hôpital's Rule allows for the differentiation of the numerator and denominator separately to determine the limit.

For instance, in the original exercise, differentiating \( x^3 \) results in \( 3x^2 \), while differentiating \( e^{2x} \) produces \( 2e^{2x} \). This process is repeated multiple times until a form that can be evaluated without indeterminacy is reached. Understanding how to correctly differentiate functions, including polynomials and exponentials, is fundamental to applying L'Hôpital's Rule effectively.

Exponential Functions

Exponential functions, like \( e^{2x} \), are crucial to mathematical analysis, particularly in limits and growth rates. They typically grow much faster than polynomial functions, which is important in understanding why certain limits, like the one in the original exercise, resolve as they do.

In this exercise, the function \( e^{2x} \) signifies exponential growth. As \( x \to \infty \), \( e^{2x} \) expands so rapidly that it eventually outweighs any polynomial grow, such as \( x^3 \). This characteristic of exponential functions is what makes the limit \( \lim _{x \rightarrow \infty} \frac{x^{3}}{e^{2x}} \) tend to zero after repeated applications of L'Hôpital's Rule. Recognizing the explosive growth nature of exponential functions helps in predicting the outcome of such limits.