Finding Horizontal Asymptote for f(x)=(2x-5)/(x^2-4): Calculus Help

  • Context: Undergrad 
  • Thread starter Thread starter ashleyk
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Discussion Overview

The discussion focuses on finding the horizontal asymptote for the function f(x) = (2x - 5)/(x^2 - 4). It involves concepts from calculus, particularly limits, although some participants note that it may not strictly require calculus.

Discussion Character

  • Technical explanation, Mathematical reasoning, Homework-related

Main Points Raised

  • One participant states the function and requests help specifically for finding the horizontal asymptote after having found the vertical asymptote.
  • Another participant proposes that the horizontal asymptote is determined by evaluating the limit as x approaches infinity, concluding that it is y = 0.
  • A different participant notes that when the degree of the polynomial in the denominator is greater than that in the numerator, the horizontal asymptote is at y = 0.
  • Another contribution suggests a method of multiplying the numerator and denominator by the inverse of x raised to the largest power of the denominator to evaluate the limit.
  • One participant describes dividing both the numerator and denominator by (1/x^2) and concludes that the horizontal asymptote is y = 0 based on the behavior of the numerator and denominator as x approaches infinity.

Areas of Agreement / Disagreement

Participants generally agree that the horizontal asymptote is y = 0, based on the reasoning that the degree of the denominator is greater than that of the numerator. However, the discussion includes different methods for arriving at this conclusion.

Contextual Notes

Some participants provide different approaches to finding the horizontal asymptote, which may imply varying levels of understanding or preference for methods. The discussion does not resolve any potential nuances in the application of limits or the definitions of asymptotes.

ashleyk
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Let f(x)= (2x-5)/(x^2-4)

I need help on finding the horizontal asymptote. I have already found the vertical. (I realize this doesn't really involve calculus but I already did other part of the problem that involved a derivative)
 
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the horizontal asymptote is

\lim_{x \rightarrow \infty} \frac{2x - 5}{x^2 -4} = 0

The horizontal asymptote is

y = 0
 
when the power in the denominator is larger than the numerator, you have a horizontal asymptote at y = 0.
 
The general idea is to multiply the numerator and denominator by the inverse of x raised to the largest power of the denominator. Then, evaluate the limit.
 
ya divide the numeraator & denominator by (1/x^2) & get that the numerator --> 0 as x --> infinity, and the denominator --> 1 or something. so the horizontal asymptote is y=0
 

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