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Magazine Chapter 4: Problem 38
Find the limit. Use I'Hopital's rule if it applies. $$\lim _{x \rightarrow \infty} \frac{x^{2}-4}{3 x^{2}+5 x+1}$$
Short Answer
Expert verified
The limit is \(\frac{1}{3}\).
Step by step solution
01
Check Limit Form
First, let's evaluate the limit form as \(x\) approaches infinity:\[ \lim _{x \rightarrow \infty} \frac{x^{2}-4}{3 x^{2}+5 x+1}\]Both the numerator and the denominator are polynomials of degree 2. When \(x\) approaches infinity, they both tend towards infinity, suggesting the \(\frac{\infty}{\infty}\) indeterminate form.
02
Apply L'Hopital's Rule
Since we have an indeterminate form \(\frac{\infty}{\infty}\), we can apply L'Hopital's Rule. This rule states that for limits of this form, we can take the derivative of the numerator and the derivative of the denominator:\[\lim _{x \rightarrow \infty} \frac{d}{dx}(x^{2}-4) \quad \, \bigg/ \quad \, \frac{d}{dx}(3 x^{2}+5 x+1) = \lim _{x \rightarrow \infty} \frac{2x}{6x+5}\]
03
Simplify the New Limit Form
Now evaluate the new limit form:\[\lim _{x \rightarrow \infty} \frac{2x}{6x+5}\] As \(x\) approaches infinity, the terms \(6x\) and \(5\) show that the dominant terms are \(2x\) in the numerator and \(6x\) in the denominator. The fraction simplifies to:\[\lim _{x \rightarrow \infty} \frac{2x}{6x} = \frac{2}{6} = \frac{1}{3}\]
04
Conclusion
The limit of the original function is \(\frac{1}{3}\). Therefore, we conclude that:\[\lim _{x \rightarrow \infty} \frac{x^{2}-4}{3 x^{2}+5 x+1} = \frac{1}{3}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
When exploring limits, we often come across expressions that don't have a clear value directly. These are called indeterminate forms. One of the most common indeterminate forms occurs when both the numerator and the denominator of a fraction tend to infinity, specifically denoted as \( \frac{\infty}{\infty} \).
- In such scenarios, the limit cannot be determined by straightforward substitution.
- L'Hopital's Rule becomes our powerful tool of choice to resolve this ambiguity by guiding us to differentiate the numerator and denominator.
Limits at Infinity
Limits at infinity describe the behavior of a function as its input grows larger without bound. These limits give insights into the end behavior of functions, which is particularly useful for graph sketching and analysis.
- When we say \( \lim_{x \rightarrow \infty} f(x) \), we are asking what value the function \( f(x) \) approaches as \( x \) becomes very large.
- Comparing the growth rates of polynomial terms, especially the ones with the highest degree, is essential to determining the behavior at infinity.
Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. When we apply L'Hopital鈥檚 rule to an indeterminate form, we differentiate both the numerator and the denominator.
- The derivative of a function gives us the rate at which the function's value is changing at any given point.
- For polynomial functions, finding derivatives is straightforward, as it involves applying the power rule: \( \frac{d}{dx}(ax^n) = nax^{n-1} \).
Polynomial Functions
Polynomials are expressions consisting of variables and coefficients combined using addition, subtraction, multiplication, and non-negative integer exponents. They can be as simple as a constant or as complex as an equation involving multiple terms.
- The degree of a polynomial is the highest power of the variable it contains. This degree dictates the end behavior of the function as \( x \to \infty \) or \( x \to -\infty \).
- In the given exercise, both the numerator and denominator are second-degree polynomials (with a leading term \( x^2 \)).
