# Is there an asymptote in this graph?

In the 2nd graph, is there an asymptote?

Thanks!

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There is a vertical asymptote $x=2$ in both pictures. –  Brian M. Scott Oct 1 '12 at 6:39
are you sure? but there is a solid black point at x=2 in the second graph o-o –  user1561559 Oct 1 '12 at 6:43
There's still an asymptote. Solid black point notwithstanding. Also, when someone goes to the trouble of giving you an answer (and a good answer too), don't you think it's a bit rude to ask if he's sure about it? –  user22805 Oct 1 '12 at 6:46
The fact that the function has a defined value at $x=2$ doesn’t keep $x=2$ from being a vertical asymptote as $x\to 2^-$. Note that $x=2$ is not a vertical asymptote as $x\to 2^+$ in either picture, though for different reasons: in the first picture $x$ can’t approach $2$ from the right. –  Brian M. Scott Oct 1 '12 at 6:47
It’s okay; I wasn’t offended. In fact it was helpful, since it pinned down what you were unclear about. –  Brian M. Scott Oct 1 '12 at 6:48

The line $x = c$ is a vertical asymptote of the graph of $y = f(x)$ if any of the following four statements is true.
1. $\lim\limits_{x \to c^+} f(x) = \infty$
2. $\lim\limits_{x \to c^+} f(x) = -\infty$
3. $\lim\limits_{x \to c^-} f(x) = \infty$
4. $\lim\limits_{x \to c^-} f(x) = -\infty$
That's it. In both graphs you show, we have that statement 3, where $c = 2$, from the definition is true. Since one of those statements is true, the definition says that $x = 2$ is a vertical asymptote in both cases. Notice, the definition doesn't say anything about the value of the function at $x = c$, only the behavior of the graph as $x$ approaches $c$.