Math Problem Statement
Solution
Let's go through the questions one by one, solving the limits provided in the image.
1.
To solve this limit, divide both the numerator and denominator by the highest power of , which is :
= \lim_{x \to \infty} \frac{3 - \frac{2}{x^2} - \frac{10}{x^3}}{\frac{4}{x^2} - \frac{2}{x} - 5}$$ As $$x \to \infty$$, all terms with $$x$$ in the denominator tend to 0: $$\frac{3 - 0 - 0}{0 - 0 - 5} = \frac{3}{-5} = -\frac{3}{5}$$ The correct answer is **A.** $$-\frac{3}{5}$$. --- ### 2. $$\lim_{x \to \infty} \frac{x^5 - 2x^4 + x^3 - 3x^2 + 2x - 7}{7 - 2x + 3x^2 - x^3 + 2x^4}$$ Again, divide the numerator and denominator by the highest power of $$x$$, which is $$x^5$$: $$\lim_{x \to \infty} \frac{\frac{x^5}{x^5} - \frac{2x^4}{x^5} + \frac{x^3}{x^5} - \frac{3x^2}{x^5} + \frac{2x}{x^5} - \frac{7}{x^5}}{\frac{7}{x^5} - \frac{2x}{x^5} + \frac{3x^2}{x^5} - \frac{x^3}{x^5} + \frac{2x^4}{x^5}}$$ This simplifies to: $$\lim_{x \to \infty} \frac{1 - \frac{2}{x} + \frac{1}{x^2} - \frac{3}{x^3} + \frac{2}{x^4} - \frac{7}{x^5}}{\frac{7}{x^5} - \frac{2}{x^4} + \frac{3}{x^3} - \frac{1}{x^2} + \frac{2}{x}}$$ As $$x \to \infty$$, the terms with powers of $$x$$ in the denominator vanish, leaving: $$\frac{1}{0} \rightarrow \infty$$ The correct answer is **E.** $$\infty$$. --- ### 3. $$\lim_{x \to \infty} \frac{(1 - 2x)^3}{(x - 1)(2x^2 + x + 1)}$$ First, expand the numerator and denominator. The numerator: $$(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3$$ The denominator: $$(x - 1)(2x^2 + x + 1) = x(2x^2 + x + 1) - 1(2x^2 + x + 1) = 2x^3 + x^2 + x - 2x^2 - x - 1 = 2x^3 - x^2$$ Now, divide both numerator and denominator by $$x^3$$: $$\lim_{x \to \infty} \frac{\frac{1}{x^3} - \frac{6}{x^2} + \frac{12}{x} - 8}{2 - \frac{1}{x}}$$ As $$x \to \infty$$, this becomes: $$\frac{-8}{2} = -4$$ The correct answer is **B.** $$-4$$. --- ### 4. $$\lim_{x \to \infty} \frac{(3x - 2)^3}{(4x + 2)^3}$$ First, expand both the numerator and the denominator: The numerator: $$(3x - 2)^3 = 27x^3 - 54x^2 + 36x - 8$$ The denominator: $$(4x + 2)^3 = 64x^3 + 96x^2 + 48x + 8$$ Now, divide both numerator and denominator by $$x^3$$: $$\lim_{x \to \infty} \frac{\frac{27x^3}{x^3} - \frac{54x^2}{x^3} + \frac{36x}{x^3} - \frac{8}{x^3}}{\frac{64x^3}{x^3} + \frac{96x^2}{x^3} + \frac{48x}{x^3} + \frac{8}{x^3}} = \lim_{x \to \infty} \frac{27 - \frac{54}{x} + \frac{36}{x^2} - \frac{8}{x^3}}{64 + \frac{96}{x} + \frac{48}{x^2} + \frac{8}{x^3}}$$ As $$x \to \infty$$, this becomes: $$\frac{27}{64}$$ The correct answer is **B.** $$\frac{27}{64}$$. --- Would you like further details on any of these steps, or do you have any questions? Here are 5 related questions you might find useful: 1. What are the key techniques for solving limits at infinity? 2. How do you handle indeterminate forms in limits? 3. What is the role of L'Hopital's Rule in solving limits involving infinity? 4. How do polynomials behave as $$x$$ approaches infinity? 5. When should you apply factorization techniques in solving limits? **Tip:** Always remember to divide by the highest power of $$x$$ in rational functions when taking limits as $$x \to \infty$$.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Limits at Infinity
Rational Functions
Polynomials
Formulas
Limit of rational functions: lim(x→∞) (ax^n + ...)/(bx^n + ...)
Factoring and simplifying polynomials
Limit as x approaches infinity for dominant term analysis
Theorems
Dominance of the highest degree term in polynomial functions
Limits of rational functions
Asymptotic behavior of polynomials
Suitable Grade Level
Grade 11-12, College Level
Related Recommendation
Solving Limits for Polynomials and Rational Functions
Limit Problems: Verifying Solutions for Limit as x Approaches Infinity
Limit Calculations with Polynomial and Rational Functions
Calculating Limits of Sequences with Rational Functions and Polynomials
Evaluate Limits with Polynomial and Rational Functions - Infinity and Undefined Forms