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Finding Asymptotes to Curve

  1. Dec 10, 2013 #1
    I am confused with solving for horizontal asymptotes. I know you are supposed to find limits to positive and negative infinity. I am able to solve for positive infinity but how are you supposed to do it for negative infinity, since you are not actually plugging in a particular value. This problem sounds like a silly question but it's really confusing me.

    An example would be (4x)/(((x^4)+1)^(1/4))

    I know the limit at positive infinity is 4, but why is it -4 at negative infinity?

  2. jcsd
  3. Dec 10, 2013 #2
    The standard interpretation of a limit as x approaches negative infinity is that we examine the behavior of the function for extremely negative, but finite, values of x. Your book will have the rigorous definition. As with any limit, we try to factor and manipulate the expression we have until we find a familiar limit or collection of limits. However, less rigorously, we can usually see what happens by using a sample extremely negative value. It is important to choose a value that is so large in absolute magnitude that none of the numbers involved in the function's expression are anywhere close to it.
    For example, the expression x^4 + 1 is very close to simply being x^4 when x is a very negative number, such as x = -(10^100). That +1 may as well be invisible (Think of what value a finite precision calculator or measurement would show). That makes the denominator very close to being simply |x|. It is positive because an even power of any number is positive. In turn, this makes your fraction very close to 4x/|x|. Since x is negative, x/|x| = -1, so we have a number very close to 4*(-1) = -4. This series of deductions can of course be made rigorous with the proper algebraic manipulation of the original expression, but it is a good idea to practice making inferences by using approximations first.
    Last edited: Dec 10, 2013
  4. Dec 10, 2013 #3


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    gene, the intuitive argument you can use on x going to infinity and minus infinity will be the same and is the best way to solve these problems (see slider's post for a full explanation). Another option however is to use
    [tex] \lim_{x\to -\infty} f(x) = \lim_{x\to \infty} f(-x) [/tex]

    which turns it into a limit you claim you know how to calculate. This isn't the best simply because it is a crutch to help you solve problems that you don't fully understand.
  5. Dec 10, 2013 #4
    ok I think I figured it out thanks!
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