In Exercises 77–80, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.)
lim x → ±∞ k(x) = 1, lim x → 1⁻ k(x) = ∞, and lim x → 1⁺ k(x) = −∞
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Identify the type of function that can have different behaviors at different points. A rational function is a good candidate because it can have vertical asymptotes and horizontal asymptotes.
To satisfy lim x → ±∞ k(x) = 1, consider a rational function where the degrees of the numerator and denominator are the same. For example, k(x) = (ax + b)/(cx + d) where a/c = 1.
To satisfy lim x → 1⁻ k(x) = ∞, the function must have a vertical asymptote at x = 1. This can be achieved by having a factor of (x - 1) in the denominator.
To satisfy lim x → 1⁺ k(x) = −∞, the function should approach negative infinity from the right of x = 1. This can be achieved by ensuring the factor (x - 1) in the denominator does not cancel with a similar factor in the numerator.
Combine these insights to construct a function like k(x) = (x + 1)/(x - 1). This function has a horizontal asymptote at y = 1, a vertical asymptote at x = 1, and the behavior around x = 1 matches the given conditions.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits at Infinity
Limits at infinity describe the behavior of a function as the input approaches positive or negative infinity. In this context, lim x → ±∞ k(x) = 1 indicates that as x becomes very large or very small, the function k(x) approaches the value 1. This suggests a horizontal asymptote at y = 1, which is crucial for sketching the graph of the function.
A vertical asymptote occurs when a function approaches infinity or negative infinity as the input approaches a specific value. Here, lim x → 1⁻ k(x) = ∞ and lim x → 1⁺ k(x) = −∞ indicate a vertical asymptote at x = 1. This means the function k(x) increases without bound as x approaches 1 from the left and decreases without bound as x approaches 1 from the right.
Piecewise functions are defined by different expressions over different intervals of the domain. In this problem, using a piecewise function can help construct a function that meets the given conditions, such as having different behaviors around x = 1 and as x approaches infinity. This flexibility is essential for creating a function that satisfies all the specified limits.