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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?

Berci
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user2041143
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2 Answers2

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

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 $x = -2$

amWhy
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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 07 '13 at 00:57
  • 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 07 '13 at 01:02
  • 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 07 '13 at 01:04
  • I already did to both. :P – user2041143 Mar 07 '13 at 01:08
  • This ain't the first time happening, but I hate when stackoverflow disallows me to select multiple answers, I have however given points to both! I appreciate anyone's help and will give as many as I can! – user2041143 Mar 07 '13 at 01:10
  • I have a last question. Why do we use the formula, n=lim x->infinity ( f(x) - k*x )? What do we determinate by it? Oh , btw k is the oblique asymptote. – user2041143 Mar 07 '13 at 01:21
  • I'd love to know why is that formula for, because it seems the formula appears every time we search for oblique. – user2041143 Mar 07 '13 at 01:30
  • n is the y-intercept value for the equation of the line of the horizontal asymptote given by the equation $y = kx + n$: where $k=\lim_{x\rightarrow+\infty}f(x)/x;; n=\lim_{x\rightarrow+\infty}(f(x)-kx)$ – amWhy Mar 07 '13 at 01:34
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    See this entry at Wikipedia: asymptotes – amWhy Mar 07 '13 at 01:35
2
  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 $\lim_{x\to\pm\infty}\frac{f(x)}{x}$.
Berci
  • 90,745
  • 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 07 '13 at 00:53
  • 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 07 '13 at 00:55
  • Check out also the other answer. Note that you still need to find the constant term in the equation of oblique asymptotes. – Berci Mar 07 '13 at 00:59
  • 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 07 '13 at 01:02