Problem 22
Question
Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty}\frac{x^2}{\sqrt{x^4 + 1}} \)
Step-by-Step Solution
Verified Answer
The limit is 1.
1Step 1: Rewrite the Expression
We start by rewriting the limit expression as \( \lim_{x \to \infty} \frac{x^2}{\sqrt{x^4 + 1}} \). This fraction is a rational function, which we can simplify by factoring within the square root.
2Step 2: Factor the Square Root
Notice that we can factor out \( x^4 \) from the square root in the denominator: \( \sqrt{x^4 + 1} = \sqrt{x^4(1 + \frac{1}{x^4})} = x^2\sqrt{1 + \frac{1}{x^4}} \). This simplifies our limit to: \( \lim_{x \to \infty} \frac{x^2}{x^2 \sqrt{1 + \frac{1}{x^4}}} \).
3Step 3: Simplify the Expression
Cancel out \( x^2 \) from the numerator and the denominator: \( \lim_{x \to \infty} \frac{x^2}{x^2 \sqrt{1 + \frac{1}{x^4}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x^4}}} \).
4Step 4: Evaluate the Limit
As \( x \to \infty \), the term \( \frac{1}{x^4} \to 0 \). Therefore, the expression inside the square root becomes \( \sqrt{1 + 0} = 1 \). So, the limit is \( \lim_{x \to \infty} \frac{1}{1} = 1 \).
Key Concepts
Rational FunctionsAsymptotic BehaviorFactorizationSquare Roots
Rational Functions
Rational functions are expressions that involve the ratio of two polynomials. They are of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. In our exercise, the rational function is \( \frac{x^2}{\sqrt{x^4 + 1}} \). Here, the polynomial in the numerator is \( x^2 \) and the denominator contains a square root expression with polynomials involved.
Understanding rational functions is essential because they often involve limits and asymptotic behavior. They can be simplified using algebraic techniques like factorization, which can make finding limits more straightforward. Notice how the denominator \( \sqrt{x^4 + 1} \) is factored to reveal simpler expressions for higher value calculations. This factorization is critical when evaluating limits of rational functions as they approach infinity.
Understanding rational functions is essential because they often involve limits and asymptotic behavior. They can be simplified using algebraic techniques like factorization, which can make finding limits more straightforward. Notice how the denominator \( \sqrt{x^4 + 1} \) is factored to reveal simpler expressions for higher value calculations. This factorization is critical when evaluating limits of rational functions as they approach infinity.
Asymptotic Behavior
Asymptotic behavior refers to the behavior of a function as the input either increases or decreases without bound. It's a way to describe how functions behave at their extremes.
In our problem, we're particularly interested in the limit as \( x \to \infty \). The rational function \( \frac{x^2}{\sqrt{x^4 + 1}} \) simplifies as \( x \) becomes very large. By factorizing, it becomes \( \frac{1}{\sqrt{1 + \frac{1}{x^4}}} \).
In our problem, we're particularly interested in the limit as \( x \to \infty \). The rational function \( \frac{x^2}{\sqrt{x^4 + 1}} \) simplifies as \( x \) becomes very large. By factorizing, it becomes \( \frac{1}{\sqrt{1 + \frac{1}{x^4}}} \).
- At \( x \rightarrow \infty \), \( \frac{1}{x^4} \rightarrow 0 \).
- The expression becomes \( \sqrt{1 + 0} = 1 \).
Factorization
Factorization is a technique used to simplify expressions or solve equations by expressing a number or an expression as a product of its factors.
In our original problem, we employed factorization to simplify \( \sqrt{x^4 + 1} \). This was rewritten as \( \sqrt{x^4(1 + \frac{1}{x^4})} = x^2\sqrt{1 + \frac{1}{x^4}} \).
In our original problem, we employed factorization to simplify \( \sqrt{x^4 + 1} \). This was rewritten as \( \sqrt{x^4(1 + \frac{1}{x^4})} = x^2\sqrt{1 + \frac{1}{x^4}} \).
- The factor \( x^4 \) is taken out to simplify the expression under the square root.
- This allows us to cancel matching terms in both numerator and denominator.
Square Roots
Square roots are often encountered in calculus, especially dealing with polynomials. A square root \( \sqrt{x} \) refers to the value that, multiplied by itself, produces \( x \).
In the problem, the denominator involves \( \sqrt{x^4 + 1} \). Understanding how to work with square roots is essential:
In the problem, the denominator involves \( \sqrt{x^4 + 1} \). Understanding how to work with square roots is essential:
- Square roots distribute over multiplication, which helps in factorization.
- It's crucial to simplify expressions like \( \sqrt{x^4 + 1} \) to \( x^2 \sqrt{1 + \frac{1}{x^4}} \) to more easily cancel terms and find limits.
Other exercises in this chapter
Problem 22
Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( f(x) = mx + b \)
View solution Problem 22
If the tangent line to \( y = f(x) \) at \( (4, 3) \) passes through the point \( (0, 2) \), find \( f(4) \) and \( f'(4) \).
View solution Problem 22
Explain why the function is discontinuous at the given number \( a \). Sketch the graph of the function. \( f(x) = \left\\{ \begin{array}{ll} \dfrac{2x^2 - 5x -
View solution Problem 22
Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to -1.5}\frac{9 - 4x^2}{3 + 2x} = 6 \)
View solution