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Function end behaviour

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the end behaviour of the function ƒ(x) = √(x2+x) - x

    2. Relevant equations
    Not quite sure, but I suppose f'(x) would be useful?

    3. The attempt at a solution
    Essentially what I did was take the derivative of f(x), however I was unsure as to how one would take the limit of the derivative as x approaches positive or negative infinity (to see if f(x) has any asymptotes). Thanks in advance for any help, sorry if I explained this poorly.
     
  2. jcsd
  3. Oct 13, 2016 #2

    BvU

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    Hello MM, :welcome:

    Thanks for introducing this "end behaviour" term to me; I didn't know about it :nb) .
    I suppose you are interested in limits to ##\pm\infty## .

    Not exactly an equation, is it ? What does it give you when ##x\rightarrow +\infty ## ?

    I suppose you are interested in limits to ##\pm\infty## .
    For ##+\infty##, how can you rewrite f(x) ?
     
  4. Oct 13, 2016 #3

    Ray Vickson

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    $$\sqrt{x^2+x}-x = \sqrt{x^2} \left(1 + \frac{1}{x} \right)^{1/2} -x = |x| \left(1 + \frac{1}{x} \right)^{1/2} -x $$
    What happens to ##(1+z)^{1/2}## for small ##z = 1/x##?
     
  5. Oct 13, 2016 #4

    Mark44

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    Not really. For the end behavior, all you need to do is to evaluate the limits ##\lim_{x \to \infty}f(x)## and ##\lim_{x \to -\infty}f(x)##.
     
  6. Oct 13, 2016 #5
    So as x approaches infinity, will (1+1/x)^(1/2) approach 1? Thus, the function approaches |x| - x ?
     
  7. Oct 13, 2016 #6

    Ray Vickson

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    Yes, as ##x \to +\infty## we have ##(1+ 1/x)^{1/2} \to 1##, but for large ##x>0## (not yet at ##x = \infty##) there will be some small correction terms; that is,
    $$\left(1 + \frac{1}{x}\right)^{1/2} = 1 + \text{small correction terms}$$
    for large ##x##. Don't forget that we multiply ##(1 + 1/x)^{1/2}## by ##|x|##, so we multiply the small correction terms by the large value of ##|x|##. The result of doing that will be something that does NOT go to zero as ##x## goes to ##\infty##. You need to figure out what the main part of the small correction term actually is, in order to arrive at a correct limiting behavior. If you did not do that you would arrive at ##\lim_{x \to \infty} \: x \cdot 1 - x = 0##, and that is not the right answer.
     
  8. Oct 13, 2016 #7
    Ah, I see. That makes sense although I am not sure how to actually go about doing that; my knowledge of calculus is very very limited. I think my teacher is showing us his solution soon so hopefully it will become clear. Thanks so much for the help! :smile:
     
  9. Oct 14, 2016 #8

    BvU

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    If you want to do some reconnoitering: Do sqrt(1.01) , sqrt(1.001) sqrt(1.0001) etc on your calculator. Or peek at the Taylor series for ##\ \sqrt{1+\epsilon}\ \ \ ##.
     
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