Is there a horizontal asymptote in $y =\frac{ x}{x^2 + 1}$

Question

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

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.

Answer 1

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.

Answer 2

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 $\infty$.