Problem 441
Question
$$ \text { If } \lim _{x \rightarrow \infty} \frac{x^{2}-1}{x+1}-a x-b=2, \text { find the values of } a \text { and } b \text { . } $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(a=1\) and \(b=-3\).
1Step 1: Expressing the given expression in a simpler form
The first step is to express the given expression \(\frac{x^{2}-1}{x+1}-a x-b-2=0\) in a simpler form. This can be done by factoring the numerator: \(x^{2}-1=(x-1)(x+1)\), so \(\frac{(x-1)(x+1)}{x+1}-a x-b-2=0\). You have \(x+1\) in both the numerator and denominator, so you can cancel them out to obtain \(x-1 - a x - b - 2 = 0 \) or \( (1-a)x - b - 3 = 0\).
2Step 2: Applying properties of limits
In the limit as \(x\) approaches infinity, the expression \( (1-a)x - b - 3 \) must be constant. Otherwise, the limit will either be negative or positive infinity, not a numerical value. So \(1-a\) should be equal to 0, otherwise the \(x \to \infty\) would make the expression infinite. Thus \(a=1\). The remaining equation is \(-b - 3 = 0\). From this, it's clear that \(b = -3\).
3Step 3: Substituting the values back
After finding the values of \(a\) and \(b\), substitute the values back into the original function to make sure the function still holds true. Doing this should result in the original function: \(\lim _{x \rightarrow \infty} \frac{x^{2}-1}{x+1}-x+3=2\). So the initially calculated values are correct.
Key Concepts
Limit at InfinityPolynomial DivisionAsymptotic Behavior
Limit at Infinity
Understanding the concept of a limit at infinity is crucial when analyzing the behavior of functions as they extend towards infinity. A limit at infinity helps us determine what value a function approaches as the input becomes very large or very negative.
In the problem, we have a function in the form of \( \lim _{x \rightarrow \infty} \frac{x^{2}-1}{x+1}-a x-b=2 \). As \( x \to \infty \), our goal is to simplify the expression so that we can find \(a\) and \(b\) which make the limit equal to 2.
We often look at the highest degree terms when \( x \to \infty \) because they define the leading behavior of the function. If the limit results in a finite number, it means the function's growth rate is sustainably balanced. Hence, resolving the terms dominantly influences how we find \(a\) and \(b\). It’s like focusing on the big roller coaster hills rather than the minor bumps and jumps in a track.
In the problem, we have a function in the form of \( \lim _{x \rightarrow \infty} \frac{x^{2}-1}{x+1}-a x-b=2 \). As \( x \to \infty \), our goal is to simplify the expression so that we can find \(a\) and \(b\) which make the limit equal to 2.
We often look at the highest degree terms when \( x \to \infty \) because they define the leading behavior of the function. If the limit results in a finite number, it means the function's growth rate is sustainably balanced. Hence, resolving the terms dominantly influences how we find \(a\) and \(b\). It’s like focusing on the big roller coaster hills rather than the minor bumps and jumps in a track.
Polynomial Division
Polynomial division is an essential operation in calculus. It’s like regular long division, but performed on polynomials instead of numbers. This technique can help simplify complex rational expressions.
To simplify the expression \( \frac{x^{2}-1}{x+1} \), we first factor the numerator as \((x-1)(x+1)\). By canceling like terms, we reduce fractions quickly. Here, \(x + 1\) cancels out leaving \( x - 1 \).
When talked through polynomial division or simplification, it makes handling limits at infinity simpler. In many calculus problems, this simplification helps us focus on leading coefficients which significantly shape the function's behavior.
To simplify the expression \( \frac{x^{2}-1}{x+1} \), we first factor the numerator as \((x-1)(x+1)\). By canceling like terms, we reduce fractions quickly. Here, \(x + 1\) cancels out leaving \( x - 1 \).
When talked through polynomial division or simplification, it makes handling limits at infinity simpler. In many calculus problems, this simplification helps us focus on leading coefficients which significantly shape the function's behavior.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as it heads towards infinity or approaches particular points. It’s about understanding if a function levels off, heads straight up, or oscillates wildly as inputs get extremely large.
When the expression \((1-a)x - b - 3 = 0\) is derived, it must be evaluated at infinity. By setting \(1-a=0\), we assure that increasing \(x\) won’t lead to an undefined or infinite solution. That’s aligning the asymptotic behavior with the requirement \( \lim _{x \rightarrow \infty} = 2 \).
By identifying and working with asymptotic behavior, we ensure the function extends in a manageable way. This understanding directly guided our determination that \(a = 1\) and \(b = -3\), balancing the equation to meet the limit condition effectively.
When the expression \((1-a)x - b - 3 = 0\) is derived, it must be evaluated at infinity. By setting \(1-a=0\), we assure that increasing \(x\) won’t lead to an undefined or infinite solution. That’s aligning the asymptotic behavior with the requirement \( \lim _{x \rightarrow \infty} = 2 \).
By identifying and working with asymptotic behavior, we ensure the function extends in a manageable way. This understanding directly guided our determination that \(a = 1\) and \(b = -3\), balancing the equation to meet the limit condition effectively.
Other exercises in this chapter
Problem 439
$$ \text { Find the constants } \left.a \text { and } b \text { such that } \lim _{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-a x-b=0 \quad \text { Ans. } a=1, b
View solution Problem 440
$$ \text { If } \lim _{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-a x-b=\infty, \text { find the value of } a \text { and } b \text { . } $$
View solution Problem 442
$$ \text { If } \lim _{x \rightarrow \infty} \frac{x^{3}+1}{x^{2}+1}-a x-b=2 \text { , then find } a \text { and } b \text { . } $$
View solution Problem 443
$$ \text { Find the constants } \left.a \text { and } b \text { such that } \lim _{x \rightarrow-\infty} \sqrt{x^{2}-x+1}-a x-b=0 \text { \\{ Ans. } a=-1, b=\fr
View solution