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$$y = \frac{x}{x^2 + 1}$$

I was trying to sketch the graph of the above function but have no idea how to draw the rest after drawing the two concaves.

I saw the graph of this function from a graphing calculator and it looks like there is an asymptote.

P.S - I have no idea how horizontal asymptotes occur.

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    $\begingroup$ When $x$ is large, $y\approx \pm\frac 1x$ $\endgroup$ – Claude Leibovici Jun 21 '15 at 11:23
  • $\begingroup$ @ClaudeLeibovici: When $x$ is large, $y \approx 0$, which answers the OP's question. $\endgroup$ – Alex M. Jun 21 '15 at 11:26
  • $\begingroup$ Horizontal asymptotes occur when the denominator has at least the same degree as the numerator. If the degrees are the same, then the asymptote is not zero. If the denominator is of higher degree than the numerator, then the asymptote is $y=0$. $\endgroup$ – Arthur Jun 21 '15 at 11:28
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    $\begingroup$ @AlexM.. This is exactly what I wanted the OP to understand in order to get the conclusion. Cheers :-) $\endgroup$ – Claude Leibovici Jun 21 '15 at 11:29
  • $\begingroup$ If $|x|>0$, then we have no problem seeing that $y = \frac{x/x}{x^2 / x + 1/ x} = \frac{1}{x + 1/x}$. Thus, for large $x$, $y \approx \frac{1}{x + 0}$, and so $y$ has a horizontal asymptote. $\endgroup$ – izœc Jun 21 '15 at 11:29
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Horizontal asymptotes for the graph of a function $f$ occur when $f(x)$ has a finite limit $\ell$ when $x$ tends to $+\infty$ or $-\infty$.

Here $\,\lim\limits_{x\to\pm\infty}\dfrac x{x^2+1}=\lim\limits_{x\to\pm\infty}\dfrac x{x^2}=0^+$ or $0^-$. Hence the $x$-axis is a horizontal asymptote to the graph. We even can say the graph is above its asymptote for $x>0$ large enough, under its asymptote for $x<0$, $\lvert x\rvert$ large enough.

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If your function has limits $\lim \limits _{x \to -\infty} f(x) = l_1$ and/or $\lim \limits _{x \to \infty} f(x) = l_2$, then the lines $y=l_1$ and $y=l_2$ are horizontal aymptotes toward $-\infty$ and, $\infty$, respectively. In your case, both limits are $0$ so the $Ox$ axis is a horizontal asymptote towards both $- \infty$ and $\infty$.

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