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Ok, well I would like to ask a few questions.

Let's say we have a function $y=\displaystyle\frac{2 x^2}{x+2}$.

So, the $D_f$ would be $\Bbb R \setminus\{-2\}$ right.

To find the asymptotes, for vertical asymptote, do we search it if there is no $D_f$ limitation (only $\Bbb R$?). And is -2 the vertical asymptote, the same as $D_f$?

How do we find horizontal asymptote, and the obliques?

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By Df, I mean the allowed values of x. And R is real numbers. –  user2041143 Mar 7 '13 at 0:48

2 Answers 2

up vote 6 down vote accepted

$$f(x) = \dfrac {2x^2}{(x + 2)}$$

The domain of the function is $\mathbb R \setminus \{-2\}$. Yes, you're correct that there is a vertical asymptote at $x = -2$. Note, as the $\lim_{x \to -2^+} f(x) = +\infty$, and $\lim_{x \to -2^{-}} f(x) = -\infty$

You can check for horizontal asymptotes by determining if/when $\lim_{x\to \pm \infty} f(x)= \pm \infty$ (if yes, exists, if no, then not).

To check for oblique asymptotes, check the limiting behavior of $\dfrac{f(x)}{x}$ as $x \to \pm \infty$. If it evaluates to $k$ then the slope of the line of the asymptote is $k$.

It helps in problems like this, as in all problems, to graph the function:

enter image description here

Note that the range of your function is $y\leq -16,\;\;y\geq 0$ Note also the vertical asymptote at $x = -2$

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I appreciate your help! Oh.. you missed a ) in your first function although it's not that important. :P –  user2041143 Mar 7 '13 at 0:57
    
glad to help! I'm not sure what you mean by missing a): I edited to address the domain (the allowed values of $x$), but are you referring to something else? –  amWhy Mar 7 '13 at 1:02
    
Oh, yea.. I get it now, since the x=-2, we change x=-2 in the lim, therefore the right/left lim are both infinity, therefore -2 is a vertical asymptote. –  user2041143 Mar 7 '13 at 1:04
    
I already did to both. :P –  user2041143 Mar 7 '13 at 1:08
1  
See this entry at Wikipedia: asymptotes –  amWhy Mar 7 '13 at 1:35
  1. Yes, only $x=-2$ is not allowed, so the domain of $f$ is indeed $\Bbb R\setminus\{-2\}$.
  2. Yes, the vertical asymptote is where the function wants to be $\pm\infty$ (in $y$ coordinate), so in this case it is at $x=-2$. But, this is not the same as $D_f$, rather its complement.
  3. For the horizontal asymptote (if any) check $\lim_{\pm\infty}f$ (so where $x$ wants to be $\pm\infty$).
  4. For the oblique asymptotes, their slope can be calculated by $\lim_{x\to\pm\infty}\frac{f(x)}{x}$.
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Thanks Berci, although I would like to ask a question, is it true that there is no vertical asymptote if there are no limitations in Df, and if there are limitations the limitations are actually the vertical asymptote? –  user2041143 Mar 7 '13 at 0:53
    
The first question is affirmative since then $y$ is never $\pm\infty$ (on $\Bbb R$, at least). For the second, we can just pick out elements from the domain, like in the case of $y=x/x$. This doesn't have vertical asymptote at $x=0$. –  Berci Mar 7 '13 at 0:55
    
Thanks for your answers! –  user2041143 Mar 7 '13 at 0:57
    
Check out also the other answer. Note that you still need to find the constant term in the equation of oblique asymptotes. –  Berci Mar 7 '13 at 0:59
    
Ok. However, if we have to find the vertical asymptote by lim, do we go this way. lim lim x->-2 (2x^2)/(x+2)? Once we switch places for x=-2, we get to divide by 0, which is impossible. –  user2041143 Mar 7 '13 at 1:02

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