So, horizontal asymptote of this rational function $\frac{x^{}}{x^{2}+4}$ is y = 0.
But this function's graph goes through the origin which is (0, 0).
I don't understand this. If horizontal asymptote doesn't allow any y value to be 0, how can the origin be plotted, which has the y value of 0? Is there an exception for origin? I thought the graph was never supposed to cross the asymptotes.
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$\begingroup$The usual definition is something like this: If $f$ is a function defined for all real numbers, then a horizontal line $y = c$ is a horizontal asymptote of $f$ if $$ \lim_{x \to \infty} f(x) = c\quad\text{and/or}\quad \lim_{x \to -\infty} f(x) = c. $$ It's a common misunderstanding that the graph of a function cannot cross a horizontal asymptote. For example, the function $$ f(x) = \frac{\sin x}{1 + x^{2}} $$ has $y = 0$ as horizontal asymptote, but the graph crosses the asymptote infinitely many times (at $x = n\pi$ for each integer $n$).
(It is true that the graph of a function cannot cross a vertical asymptote, but that's essentially a consequence of the vertical line test.)
$\endgroup$ 1 $\begingroup$Here's a verbal picture of what's going on...
The horizontal asymptote of a rational function describes end behavior. That is, it describes how the output of your graph, $f(x)$, behaves as your input, $x$, gets really really big, as in far from zero (as $x \rightarrow \pm \infty$). The graph itself can cross this asymptote.
This is a bit different than what a vertical asymptote describes. A vertical asymptote occurs at a discontinuity, where $x$ is not defined. That's why your graph can't cross the vertical asymptote. As your graph approaches the discontinuity, or the break in your graph, the function often explodes towards either $\pm \infty$.
$\endgroup$ $\begingroup$The graph of the function $y=\frac{x}{x^2+4}$ goes through $(0,0)$, but as $x$ goes to $\infty$ it flattens towards the point $(0,1)$.
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