Problem 25 - Calculus Single Variable

Question

In each Exercises \(25-32,\) use an algebraic manipulation to put the limit in a form which can be treated using l'Hôpital's Rule; then evaluate the limit. \(\lim _{x \rightarrow+\infty} x \cdot e^{-2 x}\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form

The expression given is \( \lim_{x \rightarrow +\infty} x \cdot e^{-2x} \). As \( x \to +\infty \), the term \( x \) approaches infinity and \( e^{-2x} \) approaches zero. Thus, we have the indeterminate form "\( \infty \cdot 0 \)."

Step 2: Express as a Quotient

Rewrite the expression as a quotient, which is more suitable for l'Hôpital's Rule. We can express the original limit as: \[ \lim_{x \rightarrow +\infty} \frac{x}{e^{2x}} \] Now, we have the form \( \frac{\infty}{\infty} \), which is suitable for l'Hôpital's Rule.

Step 3: Apply l'Hôpital's Rule

Since the limit is in the indeterminate form \( \frac{\infty}{\infty} \), we can apply l'Hôpital's Rule. Differentiate the numerator and the denominator separately: - Derivative of the numerator \( x \) is \( 1 \).- Derivative of the denominator \( e^{2x} \) is \( 2e^{2x} \).Now the limit becomes: \[ \lim_{x \rightarrow +\infty} \frac{1}{2e^{2x}} \]

Step 4: Evaluate the New Limit

As \( x \to +\infty \), \( e^{2x} \to +\infty \). Thus, \( \frac{1}{2e^{2x}} \to 0 \). Therefore:\[ \lim_{x \rightarrow +\infty} \frac{1}{2e^{2x}} = 0 \]

Key Concepts

Indeterminate Forms

When working with limits, you may encounter expressions that seem difficult to evaluate directly. These expressions are known as indeterminate forms. An indeterminate form arises when substituting values into a limit expression results in an undefined form like \(0/0\) or \(\infty - \infty\). In these cases, direct substitution doesn't provide a meaningful answer.
One common indeterminate form is "\(\infty \cdot 0\)," which occurs when you have a product where one term approaches infinity while the other approaches zero. Such expressions don't have a clear limit, which is why they are termed indeterminate. So, what should you do when faced with an indeterminate form? Often, you'll need to use algebraic manipulation to rewrite the expression in a way that allows further analysis, such as converting it into a quotient. This is where techniques like l'Hôpital's Rule become useful.

Limits at Infinity

Understanding limits at infinity involves looking at the behavior of a function as the input grows unbounded, either positively or negatively. For instance, if you're considering the limit \( \lim_{x \rightarrow +\infty} x \cdot e^{-2x} \), you want to understand how the function behaves when \(x\) gets very large.
In our case, as \(x\) approaches infinity, \(x\) alone goes to infinity, while \(e^{-2x}\) tends toward zero. This results in the indeterminate form "\(\infty \cdot 0\)," which brings us back to evaluating it in another way.
This is why tools such as converting the expression to a quotient help us manage infinity-based expressions effectively. Rewriting it as \( \lim_{x \rightarrow +\infty} \frac{x}{e^{2x}} \) transforms the original expression into the "\(\frac{\infty}{\infty}\)" form, where we can apply l'Hôpital's Rule to find the limit in a solvable form.

Algebraic Manipulation

Algebraic manipulation is your friend when it comes to simplifying complex expressions, especially to make them ready for solving using methods like l'Hôpital's Rule. When faced with a product that results in an indeterminate form, consider converting it into a fraction or a different expression that's more manageable.
For the expression \( \lim_{x \rightarrow +\infty} x \cdot e^{-2x} \), we applied algebraic manipulation to rewrite the product in the form of a quotient: \(\frac{x}{e^{2x}}\). This conversion from a product to a fraction makes it possible to apply l'Hôpital's Rule. You differentiate the numerator and the denominator separately, simplifying the evaluation of the limit significantly.
Remember, algebraic manipulation might involve factoring, expanding, or converting formats to facilitate limit processes, especially when special rules like l'Hôpital's are involved. This makes it a crucial step in solving limits, particularly when dealing with expressions that are not immediately apparent.