Problem 25
Question
The value of \(\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin (1 / x)+x^{2}}{1+|x|^{3}}\right]\) is (A) 1 (B) \(-1\) (C) 0 (D) \(\infty\)
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Analyze the expression
The given expression is \( \lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin (1 / x)+x^{2}}{1+|x|^{3}}\right] \). Since we are considering the limit as \( x \to -\infty \), \(|x| = -x\). Thus, the expression simplifies to \( \frac{x^{4} \sin (1 / x)+x^{2}}{1+(-x)^{3}} \).
2Step 2: Simplify the denominator
Notice that the denominator \( 1+|x|^3 = 1 + (-x)^3 \) can be rewritten as \(1 + (-x)^3 = 1 - x^3 \) as \( x \to -\infty \). In large negative values of \( x \), the \(-x^3\) term dominates, so the denominator approaches \(-x^3\).
3Step 3: Simplify the numerator
The numerator \( x^4 \sin(1/x) + x^2 \) consists of two terms: \( x^4 \sin(1/x) \) and \( x^2 \). As \( x \to -\infty \), \( \sin(1/x) \) approaches 0 since \( 1/x \to 0 \). Thus, \( x^4 \sin(1/x) \approx 0 \). The remaining term is \( x^2 \).
4Step 4: Evaluate the limit
Combine the simplified parts: \( \frac{x^2}{-x^3} \). This simplifies to \( -\frac{1}{x} \). As \( x \to -\infty \), \( -\frac{1}{x} \to 0 \).
5Step 5: Conclude the solution
Since all dominant terms have been considered, the initial expression evaluates to zero as \( x \to -\infty \). Thus, the limit is \( 0 \).
Key Concepts
InfinityTrigonometric FunctionsPolynomials
Infinity
Infinity can be a challenging concept, but it is crucial in calculus and mathematical limits. In this exercise, we explored the limit as \( x \to -\infty \). Infinity represents a value that is unbounded, and when we consider limits at infinity, we are interested in understanding the behavior of a function as it increases or decreases without bound.
When dealing with limits at infinity, consider these steps to simplify your calculations:
When dealing with limits at infinity, consider these steps to simplify your calculations:
- Identify the dominant terms in the numerator and the denominator. These are terms that will primarily influence the behavior of the function as \( x \) becomes very large (positive or negative).
- Evaluate the overall limit by simplifying and comparing these dominant terms.
Trigonometric Functions
Trigonometric functions such as \( \sin(x), \cos(x), \tan(x) \), etc., are common in calculus exercises. In this problem, we dealt with \( \sin\left( \frac{1}{x} \right) \), which is a fascinating expression when \( x \to -\infty \).
Here’s why it simplifies nicely:
Here’s why it simplifies nicely:
- As \( x \to -\infty \), the expression \( \frac{1}{x} \) tend towards 0 since dividing by a large negative number yields values close to zero. Consequently, \( \sin\left( \frac{1}{x} \right) \) behaves similarly.
- Because \( \sin(x) \) approximates its argument when close to zero, \( \sin\left( \frac{1}{x} \right) \approx \frac{1}{x} \). This helps simplify complex expressions and is especially useful in limits and calculi.
Polynomials
Polynomials are algebraic expressions consisting of terms made up of variables raised to a power. In our exercise, the expression contained polynomial terms like \( x^4 \) and \( x^2 \). Identifying and leveraging these terms are essential when simplifying limits at infinity.
Polynomials have properties that make them manageable when evaluating limits:
Polynomials have properties that make them manageable when evaluating limits:
- The behavior of a polynomial at infinity is primarily governed by its highest degree term. Therefore, as \( x \to -\infty \), the term with the highest degree becomes dominant.
- For instance, in our exercise, \( x^4 \sin(1/x) + x^2 \), the \( x^4 \sin(1/x) \) term simplifies down substantially as \( x \to -\infty \), primarily because \( \sin(1/x) \to 0 \), allowing \( x^2 \) to become the dominant term in the numerator.
- This principle allows us to ignore smaller order terms when determining the behavior of the polynomial at extreme values of \( x \), streamlining the solving process for limits.
Other exercises in this chapter
Problem 23
\(\lim _{x \rightarrow 1 \atop y \rightarrow 0} \frac{y^{3}}{x^{3}-y^{2}-1}\) as \((x, y) \rightarrow(1,0)\) along the line \(y=\) \(x-1\) is given by (A) 1 (B)
View solution Problem 24
\(\lim _{n \rightarrow \infty} \frac{1-2+3-4+5-6+\ldots-2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}\) is equal to (A) \(\frac{1}{3}\) (B) \(-\frac{1}{3}\) (C) \(-\fra
View solution Problem 26
\(\lim _{x \rightarrow 2} \frac{2^{x}+2^{3-x}-6}{2^{-x / 2}-2^{1-x}}\) is equal to (A) 8 (B) 4 (C) 2 (D) None of these
View solution Problem 28
\(\lim _{n \rightarrow \infty}\left[\log _{n-1}(n) \cdot \log _{n}(n+1) \cdot \log _{n+1}(n+2) \ldots \log _{n^{t}-1}\left(n^{k}\right)\right]\) is equal to (A)
View solution