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Rewrite the indeterminate form of type \(0 \cdot \infty\) as either type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty} .\) Use L'Hôpital's Rule to evaluate the limit. $$ \lim _{x \rightarrow \infty}(3 x)\left(\frac{2}{e^{x}}\right) $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the Indeterminate Form

Consider the expression \((3x)\left(\frac{2}{e^{x}}\right)\). When \(x \to \infty\), \(3x\) tends to \(\infty\) and \(\frac{2}{e^x}\) tends to 0. This leads to the indeterminate form \(0 \cdot \infty\).
02

Rewrite as a Quotient

We need to rewrite the expression to apply L'Hôpital's Rule. Let's convert the product \((3x)\left(\frac{2}{e^x}\right)\) into a quotient: \(\frac{3x \cdot 2}{e^x} = \frac{6x}{e^x}\). This is now a form of \(\frac{\infty}{\infty}\) when \(x \to \infty\).
03

Apply L'Hôpital's Rule

Apply L'Hôpital's Rule to the limit \(\lim_{x \to \infty} \frac{6x}{e^x}\) by differentiating the numerator and the denominator. The derivative of the numerator \(6x\) is \(6\), and the derivative of the denominator \(e^x\) is \(e^x\). The limit can be rewritten as \(\lim_{x \to \infty} \frac{6}{e^x}\).
04

Evaluate the Limit

Now evaluate \(\lim_{x \to \infty} \frac{6}{e^x}\). As \(x\) approaches infinity, the exponential \(e^x\) grows much faster than any constant, making \(\frac{6}{e^x}\) approach \(0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
In calculus, an indeterminate form arises when the limit of a function cannot be directly determined using standard methods. These forms take various shapes, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), among others. When you first encounter them, they might seem puzzling because they don't lead to an obvious conclusion about the limit. For instance, with \( 0 \cdot \infty \), one part of the function trends towards zero, suggesting the whole product should be zero. However, the other part trends towards infinity, which suggests the whole product should be very large.

To resolve indeterminate forms like \( 0 \cdot \infty \), a transformation often needs to occur — turning them into forms more standard like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These transformations allow you to utilize powerful calculus methods like L'Hôpital's Rule, which are designed to handle these tricky cases. Understanding how to identify and transform indeterminate forms is crucial for effectively solving limit problems.
Limits in Calculus
Limits in calculus are foundational as they form the backbone of many other concepts, including derivatives and integrals. A limit describes the behavior of a function as the input approaches a certain value. With limits, mathematicians can define slope for tangents to curves and area under curves, leading to practical applications in different fields.

When approaching a problem involving limits, it's essential to analyze the growth or decay of components within the function. For example, in indeterminate forms like \( \lim _{x \rightarrow \infty}(3 x) \left(\frac{2}{e^{x}}\right) \), seeing how \( 3x \) grows very large while \( \frac{2}{e^x} \) shrinks to zero is key to identifying its indeterminate nature.

By rewriting the expression using algebraic transformation, we can simplify the handling of limits. Instead of tackling a product directly, we reframe it as a quotient — once translated to \( \frac{6x}{e^x} \), this becomes a more familiar form where known methods like L'Hôpital's Rule can be directly applied.
Calculus Methods
Calculus methods are strategies and approaches for solving mathematical problems related to changes and motion. They include differentiation, integration, and techniques for evaluating limits. Among these, evaluating limits using L'Hôpital's Rule stands out as a sophisticated yet essential tool.

L'Hôpital's Rule applies when evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It involves differentiating the numerator and the denominator separately, which can simplify and solve otherwise complex problems.

In our exercise, transforming the initial expression into a quotient \( \frac{6x}{e^x} \), enabled the use of L'Hôpital's Rule. Differentiating gave \( \frac{6}{e^x} \), which, when evaluated as \( x \) approaches infinity, easily showed the function tends towards 0.
  • Ensure the original problem is indeed an indeterminate form requiring L'Hôpital's.
  • Carefully differentiate separate top and bottom terms; errors here can lead to wrong conclusions.
  • Simplify after differentiation whenever possible to make evaluating the limit intuitive.

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