Problem 11
Question
The value of \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+b x+4}{x^{2}+a x+5}\right)\) is (a) \(b / a\) (b) 1 (c) 0 (d) \(4 / 5\)
Step-by-Step Solution
Verified Answer
The value of the limit is 1 (option b).
1Step 1 - Identify the Dominant Terms
When evaluating the limit as \(x\) approaches infinity, the dominant terms in the numerator and denominator are those with the highest powers. In this case, \(x^2\) is the dominant term in both the numerator \(x^2 + bx + 4\) and the denominator \(x^2 + ax + 5\).
2Step 2 - Simplify the Limit Expression
Divide every term in both the numerator and the denominator by the dominant term, \(x^2\). This gives us: \(\lim_{x \to \infty} \left(\frac{x^2/x^2 + bx/x^2 + 4/x^2}{x^2/x^2 + ax/x^2 + 5/x^2}\right)\).
3Step 3 - Evaluate the Resulting Expression
Simplify the expression: \(\lim_{x \to \infty} \left(\frac{1 + b/x + 4/x^2}{1 + a/x + 5/x^2}\right)\). As \(x\) approaches infinity, \(b/x\), \(4/x^2\), \(a/x\), and \(5/x^2\) all approach zero.
4Step 4 - Conclude the Limit Value
With the higher degree terms approaching zero, the fraction simplifies to \(\frac{1}{1}\). Therefore, the limit is 1.
Key Concepts
Dominant Terms MethodEvaluating Limits at InfinitySimplifying Rational Expressions
Dominant Terms Method
When dealing with limits, especially as variables approach infinity, the 1dominant terms method d is like focusing on the loudest in a crowd - the terms with the highest power. In polynomial expressions, these terms overshadow all others as the variable grows larger.
For example, consider the expression c\(x^2 + bx + 4\) d. As \(x\) becomes very large, \(x^2\) becomes much more significant than any \(bx\) or constant like 4. This means \(x^2\) is the dominant term. Similarly, in the denominator \(x^2 + ax + 5\), \(x^2\) is also the dominant term.
For example, consider the expression c\(x^2 + bx + 4\) d. As \(x\) becomes very large, \(x^2\) becomes much more significant than any \(bx\) or constant like 4. This means \(x^2\) is the dominant term. Similarly, in the denominator \(x^2 + ax + 5\), \(x^2\) is also the dominant term.
- This method simplifies the process of evaluating limits.
- Focuses only on terms that actually affect the limit at infinity.
Evaluating Limits at Infinity
As you tackle limits where \(x\) approaches infinity, it's crucial to simplify and observe the behavior of the expression in such conditions.
First, by identifying dominant terms, you divide each term by the highest power variable. Using our example, each term in the numerator and denominator is divided by \(x^2\), the dominant power: \[\lim_{x \to \infty} \left(\frac{x^2/x^2 + bx/x^2 + 4/x^2}{x^2/x^2 + ax/x^2 + 5/x^2}\right)\].
This results in \[1 + \frac{b}{x} + \frac{4}{x^2}\text{ over } 1 + \frac{a}{x} + \frac{5}{x^2}\].
Now, as \(x\) approaches infinity:
This showcases how dominating terms dictate the behavior at infinity, proving that simpler expressions often determine the outcome in such scenarios.
First, by identifying dominant terms, you divide each term by the highest power variable. Using our example, each term in the numerator and denominator is divided by \(x^2\), the dominant power: \[\lim_{x \to \infty} \left(\frac{x^2/x^2 + bx/x^2 + 4/x^2}{x^2/x^2 + ax/x^2 + 5/x^2}\right)\].
This results in \[1 + \frac{b}{x} + \frac{4}{x^2}\text{ over } 1 + \frac{a}{x} + \frac{5}{x^2}\].
Now, as \(x\) approaches infinity:
- \(\frac{b}{x} \rightarrow 0\)
- \(\frac{4}{x^2} \rightarrow 0\)
- \(\frac{a}{x} \rightarrow 0\)
- \(\frac{5}{x^2} \rightarrow 0\)
This showcases how dominating terms dictate the behavior at infinity, proving that simpler expressions often determine the outcome in such scenarios.
Simplifying Rational Expressions
Simplifying rational expressions involves reducing an expression to its simplest form. When evaluating limits, particularly at infinity, this step is essential. It allows us to focus on the core behavior of the function.
Consider \[\frac{x^2 + bx + 4}{x^2 + ax + 5}\].
We began by dividing every term by \(x^2\), the dominant term.
This transforms it to \[\frac{1 + b/x + 4/x^2}{1 + a/x + 5/x^2}\], which simplifies to \(1\), as discussed earlier.
Here's why simplifying is beneficial:
Consider \[\frac{x^2 + bx + 4}{x^2 + ax + 5}\].
We began by dividing every term by \(x^2\), the dominant term.
This transforms it to \[\frac{1 + b/x + 4/x^2}{1 + a/x + 5/x^2}\], which simplifies to \(1\), as discussed earlier.
Here's why simplifying is beneficial:
- It minimizes complex calculations.
- Provides a clearer view of how functions behave as variables change significantly.
- Makes it easier to predict outcomes, like limits, quickly.
Other exercises in this chapter
Problem 9
\(\lim _{n \rightarrow \infty}\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n+1)(2 n+3)}\right]\) is equal to (a) 0 (b) \(1 / 2\) (c) \(1
View solution Problem 10
\(\lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^{5}}{(2 x+3)^{45}}=\) (a) 16 (b) 24 (c) 32 (d) 8
View solution Problem 12
\(\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+8 x+3}-\sqrt{x^{2}+4 x+3}\right)\) is equal to (a) 0 (b) \(\infty\) (c) 2 (d) \(1 / 2\)
View solution Problem 13
If \(x_{n}=\frac{1-2+3-4+5-6+\ldots \ldots \ldots 2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}\), then \(\lim _{n \rightarrow \infty} x_{n}\) is equal to (a) \(1 / 3\)
View solution