Problem 22
Question
15-36 Find the limit. $$ \lim _{x \rightarrow \infty} \frac{x+2}{\sqrt{9 x^{2}+1}} $$
Step-by-Step Solution
Verified Answer
The limit is \( \frac{1}{3} \).
1Step 1: Identify the dominant terms
The given function is \( f(x) = \frac{x+2}{\sqrt{9x^2+1}} \). As \( x \to \infty \), the dominant terms in the numerator and the denominator need to be identified. In the numerator, \( x \) is the dominant term, and in the denominator, \( \sqrt{9x^2} \) is the dominant term because \( 1 \) becomes negligible for large \( x \).
2Step 2: Simplify the expression using dominant terms
Using the dominant terms, the function simplifies to \( \frac{x}{\sqrt{9x^2}} \). Simplify the expression inside the square root to get \( \sqrt{9} \cdot \sqrt{x^2} = 3|x| \). For \( x \to \infty \), the absolute value \( |x| \) is simply \( x \), so the denominator becomes \( 3x \).
3Step 3: Reduce the expression
The simplified form \( \frac{x}{3x} \) can now be reduced by cancelling out \( x \) in the numerator and the denominator, yielding \( \frac{1}{3} \).
4Step 4: Evaluate the limit
Since all transformations and cancellations maintain the inequality and validity for the limit evaluation as \( x \to \infty \), the limit \( \lim_{x \to \infty} \frac{x+2}{\sqrt{9x^2+1}} \) is equal to \( \frac{1}{3} \).
Key Concepts
Dominant Terms and Their Role in LimitsInfinity Limits and Simplification TechniquesLimits and Continuity in Function Behavior
Dominant Terms and Their Role in Limits
When finding the limit of a function as the variable approaches infinity, identifying dominant terms is crucial. In any given function, not all terms contribute equally when the variable grows large. Dominant terms are those that significantly affect the value of the function for large inputs.
In the expression \( \frac{x+2}{\sqrt{9x^2+1}} \), as \( x \) grows very large, the term \( x \) in the numerator becomes the significant player, overshadowing the constant \( 2 \). Similarly, in the denominator, \( \sqrt{9x^2} \) dominates over \( 1 \) because the presence of \( x^2 \) makes the impact of \( 1 \) negligible.
Choosing dominant terms simplifies the problem, letting us approximate the behavior of the function at extreme values. This approach gives us a streamlined expression that is easier to work with, ensuring a more straightforward calculation of the limit.
In the expression \( \frac{x+2}{\sqrt{9x^2+1}} \), as \( x \) grows very large, the term \( x \) in the numerator becomes the significant player, overshadowing the constant \( 2 \). Similarly, in the denominator, \( \sqrt{9x^2} \) dominates over \( 1 \) because the presence of \( x^2 \) makes the impact of \( 1 \) negligible.
Choosing dominant terms simplifies the problem, letting us approximate the behavior of the function at extreme values. This approach gives us a streamlined expression that is easier to work with, ensuring a more straightforward calculation of the limit.
Infinity Limits and Simplification Techniques
Infinity limits deal with how functions behave as the variable approaches an infinitely large value or grows infinitely small. For limits at infinity, our goal is to understand the asymptotic behavior of the function by simplifying the expression using dominant terms.
The simplified expression \( \frac{x}{\sqrt{9x^2}} \) is a common example, where both terms in the numerator and denominator are rewritten using dominant components. Here, we observe that \( \sqrt{9x^2} = 3|x| \). As \( x \) moves towards infinity, \( |x| \) becomes \( x \), and the expression reduces to \( \frac{x}{3x} \).
In this format, canceling terms becomes a straightforward task, as both \( x \)s can be eliminated. It simplifies dramatically to \( \frac{1}{3} \), depicting how limiting behavior at infinity focuses more on the overall shape of the function than the details of smaller terms.
The simplified expression \( \frac{x}{\sqrt{9x^2}} \) is a common example, where both terms in the numerator and denominator are rewritten using dominant components. Here, we observe that \( \sqrt{9x^2} = 3|x| \). As \( x \) moves towards infinity, \( |x| \) becomes \( x \), and the expression reduces to \( \frac{x}{3x} \).
In this format, canceling terms becomes a straightforward task, as both \( x \)s can be eliminated. It simplifies dramatically to \( \frac{1}{3} \), depicting how limiting behavior at infinity focuses more on the overall shape of the function than the details of smaller terms.
Limits and Continuity in Function Behavior
Limits and continuity are foundational concepts in calculus that help explain how functions behave as inputs change. Having a clear understanding of these notions is essential when analyzing expressions for very large or very small input values.
The limit \( \lim_{x \to \infty} \frac{x+2}{\sqrt{9x^2+1}} = \frac{1}{3} \) illustrates an aspect of continuity, where despite potential disparities at finite values, the function trends towards a stable behavior as \( x \) becomes indefinitely large. It underscores how mathematical analysis provides a consistent framework to predict outcomes, ensuring continuity isn’t disrupted as variables extend to extreme values.
This scenario shows that the function gradually levels off to a particular value, confirming that understanding limits allows us to confidently describe how functions smooth out and regulate their fluctuating behavior over vast ranges. Understanding this principle helps in fields like engineering and economics, among others.
The limit \( \lim_{x \to \infty} \frac{x+2}{\sqrt{9x^2+1}} = \frac{1}{3} \) illustrates an aspect of continuity, where despite potential disparities at finite values, the function trends towards a stable behavior as \( x \) becomes indefinitely large. It underscores how mathematical analysis provides a consistent framework to predict outcomes, ensuring continuity isn’t disrupted as variables extend to extreme values.
This scenario shows that the function gradually levels off to a particular value, confirming that understanding limits allows us to confidently describe how functions smooth out and regulate their fluctuating behavior over vast ranges. Understanding this principle helps in fields like engineering and economics, among others.
Other exercises in this chapter
Problem 22
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