Calculating asymptote of the function

  • Thread starter Zenga
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  • #1
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Homework Statement



I would like to find an asymptote of the following function:
f(x) = [itex]\sqrt{\frac{x^3}{x+1}} + x[/itex] as x goes to negative infinity.

2. The attempt at a solution

I calculated the limit of the function as x goes to -∞ which is ∞.
However, this is not enough for me. I would like to be as precise as possible when drawing a graph of the function as x goes to -∞. This is the reason why do I want to calculate the asymptote of the function as x goes to -∞
 

Answers and Replies

  • #2
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There is no asymptote as x goes to negative infinity, there is however an obvious asymptote in the vicinity of -1, can you see it?
 
  • #3
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By taking the limit as x goes to infinity, you are seeing whether there is a horizontal asymptote or not. This is pretty much as mathematically precise as you can get. Getting the result negative infinity implies that the graph does not approach any specific value as x approaches infinity--the behavior is unbounded.

Edit: You should also calculate the limit as x goes to negative infinity.
But as Vorde said, there is no horizontal asymptote, but maybe a vertical one.
 
  • #4
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Vorde, Bashyboy: I can only determine the "asymptote" as x approaches 0, which has a slope of -45° (first derivative of the function is -1 when x = 0).

But you have probably something else in mind, isn't it?
 
  • #5
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There is no asymptote at x = 0.
 
  • #6
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I agree - there is no asymptote - we can only determine the angle of the slope to the graph in x = 0.
 
  • #7
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I think you need to restate what exactly it is that you are trying to search for. In your original post, you speak about finding asymptotes and such; now you are discussing slope angles.
 
  • #8
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I am looking for an oblique asymptote.
 
  • #9
Mentallic
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A common approach would be to do the following:

[tex]y=\sqrt{\frac{x^3}{x+1}}+x[/tex]

[tex]=\frac{\sqrt{x^3}}{\sqrt{x+1}}+x[/tex]

[tex]=\frac{\sqrt{x^3}+x\sqrt{x+1}}{\sqrt{x+1}}[/tex]

But this is wrong! Because we're considering when [itex]x\to -\infty[/itex] and [tex]\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/tex] only works for positive a and b.

So what can we do? Well, we already know that even though [itex]x\to -\infty[/itex] that the value under the square root sign will be positive, so how about we let [itex]x=-x_1[/itex], substitute that into [tex]y=\sqrt{\frac{x^3}{x+1}}+x[/tex] and now we can consider the limit of [itex]x_1 \to\infty[/itex] and follow the same process as before.

After you've done that, you want to simplify the numerator by multiplying both numerator and denominator by the numerator's conjugate (in order to get rid of the surds).
 
Last edited:
  • #10
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And to point you in the right direction, I wolfram alpha'd this (god I love that site), I'll say there is an asymptote at -1, so it should be pretty easy to mathematically prove that.
 
  • #11
Mentallic
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And to point you in the right direction, I wolfram alpha'd this (god I love that site), I'll say there is an asymptote at -1, so it should be pretty easy to mathematically prove that.


We're looking for the horizontal (or oblique) asymptote as [itex]x\to -\infty[/itex] and you don't need wolfram alpha to realize that [itex]x\neq -1[/itex]. Clearly the denominator under the square root cannot be equal to 0.
 

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