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Calc I limit (find the horizontal asymptote)

  1. Jul 22, 2009 #1
    I'm a calc newb, and I am a little stumped here. Thanks for your help. How do you do this?

    http://www.webassign.net/www29/symImages/0/8/103b04681b693242466ef17cefccc1.gif
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jul 22, 2009 #2

    Office_Shredder

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    What's the limit as x goes to infinity of x5-x7? Then try to figure out the limit that you're given
     
  4. Jul 22, 2009 #3
    If the limit exists, denote it by L. Then
    [tex]
    \begin{align*}
    \lim_{x\to\infty} \arctan(x^5-x^7) &= L \\
    \lim_{x\to\infty} (x^5-x^7) &= \tan(L)
    \end{align*}
    [/tex]
    So as the left side approaches -infinity, what does L have to approach? Remember tan=sin/cos.
     
  5. Jul 22, 2009 #4
    Remember that if a function is continuous, you can take the limit 'inside'. By that, I mean that if f(x) is continuous, then [tex] \lim_{x \rightarrow \infty} f(x) = f(\lim_{x \rightarrow \infty} x) [/tex].



    arctan(x) is a continuous function, so just like Office Shredder first suggested, find the limit of x^5 - x^7 first.
     
  6. Jul 22, 2009 #5

    Office_Shredder

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    Not to spoil the surprise, but

    While the idea in the two above posts is correct, it should be noted that in this case

    [tex] \lim_{x \rightarrow \infty} (x^5-x^7)[/tex]

    is NOT in the domain of either tan(x) or arctan(x) so you technically can't arctan out of the limit, and in this case the limit is not in the domain of tan(x) either. So while doing what JG and n!k posted should give a good idea as to what the answer is, they can't be used as methods for final solutions
     
  7. Jul 22, 2009 #6
    Office_Shredder, I thought that arctan(x) was defined for all x and also continuous for all x?
     
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