- Find the limit of a sum, a difference, and a product.
- Find the limit of a polynomial.
- Find the limit of a power or a root.
- Find the limit of a quotient.
Analyzing Business Functions
Business models often use polynomial and rational functions to predict profit, cost, and revenue. Understanding limits helps us analyze what happens as production levels change, even at points where functions have discontinuities.
Question Help: Evaluating Limits Using Properties
- For polynomial functions, use direct substitution: [latex]\lim_{x \to a} p(x) = p(a)[/latex]
- For quotients that give [latex]\frac{0}{0}[/latex], try factoring and simplifying first
- For roots in quotients, consider multiplying by a conjugate
- Always try direct substitution first to see if it works
A company’s monthly profit in thousands of dollars is modeled by [latex]P(x) = 2x^3 - 3x + 1[/latex], where [latex]x[/latex] represents the number of units produced (in hundreds). As production approaches 500 units ([latex]x[/latex] approaches 5), what profit does the model predict?
Find [latex]\lim_{x \to 5} (2x^3 - 3x + 1)[/latex].
A company’s quarterly revenue in thousands of dollars is modeled by [latex]R(x) = x^4 - 4x^3 + 5[/latex], where [latex]x[/latex] represents thousands of units sold. Find [latex]\lim_{x \to -1} R(x)[/latex] to determine what the model predicts as sales approach -1 thousand units (though negative sales don’t make physical sense, the mathematical limit exists).
[latex]\lim_{x \to -1} (x^4 - 4x^3 + 5) =[/latex] [response area]
Correct answer: 10
Feedback for correct answer: Excellent! You used direct substitution: [latex]-1^4 - 4(-1)^3 + 5 = 1 - 4(-1) + 5 = 1 + 4 + 5 = 10[/latex] thousand dollars. Since this is a polynomial, the limit equals the function value at [latex]x = -1[/latex].
Feedback for incorrect answer: Since this is a polynomial function, you can find the limit by substituting [latex]x = -1[/latex] directly into the expression. Calculate [latex]-1^4 - 4(-1)^3 + 5[/latex], being careful with signs and exponents.
A company’s average cost function (in dollars per unit) is [latex]C(x) = \frac{x^2 - 6x + 8}{x - 2}[/latex] for [latex]x[/latex] hundred units produced. The function is undefined at [latex]x = 2[/latex] due to a discontinuity in the cost data. Find [latex]\lim_{x \to 2} C(x)[/latex] to determine what the average cost approaches as production nears 200 units.
Solution:
First, try direct substitution:
[latex]\frac{(2)^2 - 6(2) + 8}{2 - 2} = \frac{4 - 12 + 8}{0} = \frac{0}{0}[/latex]
This is indeterminate, so we need to factor and simplify:
[latex]\begin{aligned} \lim_{x \to 2} \frac{x^2 - 6x + 8}{x - 2} &= \lim_{x \to 2} \frac{(x - 2)(x - 4)}{x - 2} && \text{factor the numerator} \ &= \lim_{x \to 2} \frac{\cancel{(x - 2)}(x - 4)}{\cancel{x - 2}} && \text{cancel common factors} \ &= \lim_{x \to 2} (x - 4) && \text{simplify} \ &= 2 - 4 \ &= -2 \end{aligned}[/latex]
As production approaches 200 units, the average cost approaches -$2 per unit. The negative value suggests the model breaks down near this production level, but mathematically the limit exists.
A manufacturer’s efficiency ratio is given by [latex]E(x) = \frac{x^2 - 11x + 28}{7 - x}[/latex] when producing [latex]x[/latex] hundred units. Find [latex]\lim_{x \to 7} E(x)[/latex].
[latex]\lim_{x \to 7} \frac{x^2 - 11x + 28}{7 - x} =[/latex] [response area]
Correct answer: -3
Feedback for correct answer: Great work! You factored the numerator as [latex](x - 4)(x - 7)[/latex], then noticed that [latex]7 - x = -(x - 7)[/latex]. After factoring out -1 from the denominator, you could cancel [latex](x - 7)[/latex] to get [latex]\lim_{x \to 7} -(x - 4) = -(7 - 4) = -3[/latex].
Feedback for incorrect answer: Direct substitution gives [latex]\frac{0}{0}[/latex], so factor first. The numerator factors as [latex](x - 4)(x - 7)[/latex]. Notice that [latex]7 - x = -(x - 7)[/latex]. Factor out the -1, then cancel common factors before evaluating.