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Use \(I^{\prime}\) Hôpital's Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. $$\lim _{x \rightarrow \infty} \frac{2 x^{2}+3 x}{x^{3}+x+1}$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Analyze the Limit

First, observe the limit \( \lim_{x \rightarrow \infty} \frac{2x^2 + 3x}{x^3 + x + 1} \). As \( x \rightarrow \infty \), both the numerator and the denominator tend toward infinity, creating an indeterminate form \( \frac{\infty}{\infty} \). This suggests the possible use of \(I'\) Hôpital's Rule.
02

Apply I'Hôpital's Rule

Since the limit presents an indeterminate form \(\frac{\infty}{\infty}\), apply \(I'\) Hôpital's Rule by differentiating the numerator and the denominator. The derivative of the numerator \( 2x^2 + 3x \) is \( 4x + 3 \), and the derivative of the denominator \( x^3 + x + 1 \) is \( 3x^2 + 1 \).
03

Evaluate New Limit

Evaluate the new limit: \[\lim_{x \rightarrow \infty} \frac{4x + 3}{3x^2 + 1}\]As \( x \rightarrow \infty \), the degree of \( x \) in the denominator is greater than in the numerator, thus the limit simplifies to 0.
04

Evaluate Using Algebraic Simplification

Rewrite the original expression by dividing every term by \( x^3 \), which is the highest power in the denominator:\[\lim_{x \rightarrow \infty} \frac{2x^2 + 3x}{x^3 + x + 1} = \lim_{x \rightarrow \infty} \frac{\frac{2x^2}{x^3} + \frac{3x}{x^3}}{\frac{x^3}{x^3} + \frac{x}{x^3} + \frac{1}{x^3}} = \lim_{x \rightarrow \infty} \frac{\frac{2}{x} + \frac{3}{x^2}}{1 + \frac{1}{x^2} + \frac{1}{x^3}}\]As \(x \rightarrow \infty\), the terms containing \( \frac{1}{x} \), \( \frac{1}{x^2} \), and \( \frac{1}{x^3} \) all approach 0, simplifying the limit to 0.

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Key Concepts

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

Understanding Indeterminate Forms
When calculating the limit of a function as it approaches infinity or zero, you may encounter situations where the expression doesn't initially lead to a clear result. These situations are known as indeterminate forms. A common indeterminate form is \( \frac{\infty}{\infty} \), which occurs when both the numerator and the denominator approach infinity. This form does not directly suggest a specific numerical limit, hence the term 'indeterminate'.

Indeterminate forms signal that the limit isn't straightforward and requires further analysis or manipulation to evaluate. Recognizing an indeterminate form early on helps you decide which calculus techniques, such as L'Hôpital's Rule, to apply for simplifying or finding the limit.
  • Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others like \( \infty - \infty \).
  • Identifying these helps direct which method to use for evaluation.
Mastering Limit Evaluation
Limit evaluation is a fundamental technique in calculus, helping to find the behavior of functions as they approach a certain point. In the given exercise, the limit \( \lim _{x \rightarrow \infty} \frac{2 x^{2}+3 x}{x^{3}+x+1} \) presented an indeterminate form, which invited using L'Hôpital's Rule.

When applying L'Hôpital's Rule, you first check that you indeed have an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). You then differentiate the numerator and the denominator separately and re-evaluate the limit. This technique helps in simplifying expressions where algebra alone might be challenging.
  • Evaluate whether a form is indeterminate before applying any rule.
  • Differentiation lowers the polynomial degree, simplifying the function.
  • The rule can be applied repeatedly if indeterminate forms persist.
Utilizing Calculus Techniques
Calculus offers a variety of methods to analyze and solve limits beyond L'Hôpital's Rule. Sometimes, simple algebraic manipulation can simplify expressions for direct limit evaluation without needing differentiation.

In the exercise, besides L'Hôpital's Rule, you could also use algebraic simplification. By dividing each term by the highest relevant power of \( x \) in the expression, you could reduce complex fractions into simpler terms. This method reveals the terms that dominate behavior as \( x \) approaches infinity, making it easier to determine the limit without derivatives.
  • Divide the terms by the largest power of \( x \) in the expression to simplify.
  • Recognize that terms with \( x \) raised to negative powers approach zero as \( x \rightarrow \infty \).
  • Choose the method based on the function's complexity and the presence of indeterminate forms.

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