# More help with vertical asymptotes

Of the six questions regarding finding vertical asymptotes of graphs, I've had problems with two. The second is using the graph of $$g(x)= \frac{3+x}{x^{2}(3-x)}$$

Now, looking at the function, it seems that both 0 and 3 will be the vertical asymptotes, since $$g(0)= \frac{3+0}{0^{2}(3-0)}= \frac{3}{0(3)}= \frac{3}{0}= undefined$$ and $$g(3)= \frac{3+3}{3^{2}(3-3)}= \frac{6}{9(0)}= \frac{6}{0}= undefined$$

However, those answers were not accepted, why?

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Probably a more detailed analysis was asked: Observe that $\lim_{x \searrow 3} g(x) = - \infty$ and $\lim_{x \nearrow 3} g(x) = + \infty$. The picture at $0$ is different. Have a look at wolframalpha.com/input/?i=%28x%2B3%29%2F%28x%5E2+%283-x%29%29 – t.b. Feb 13 '11 at 17:31
Accepted by whom? What is the definition of vertical asymptote you are working with? – Mariano Suárez-Alvarez Feb 13 '11 at 17:38
@mariano, this calculus class homework is submitted and processed by webassign.net. The definition I'm working with is if f(x) approaches positive or negative infinity, as x approaches c from the right or left, then the line x=c is a vertical asymptote of the graph of f – Jason Feb 13 '11 at 18:32

The usual definition for a vertical asymptote at $x=a$ is that at least one of the following conditions must be satisfied.

1. $\displaystyle \lim_{x \rightarrow a^+} f(x) = \pm \infty$
2. $\displaystyle \lim_{x \rightarrow a^-} f(x) = \pm \infty$

It is clear from the equations and from the graph of the function that at least one of these conditions is satisfied at $x=0$ and $x=3$.

Hence, $x=0$ and $x=3$ are both vertical asymptotes to the function.

However, I have seen some people defining vertical asymptote as

$$\displaystyle \lim_{x \rightarrow a^-} f(x) = \displaystyle \lim_{x \rightarrow a^+} f(x) = \pm \infty$$

i.e. the left limit and the right limit must both tend to $+\infty$ or the left limit and the right limit must both tend to $-\infty$. In that case, $x=3$ won't be consider a vertical asymptote.

In general, I prefer to work with the first definition.

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and that's the same definition as provided in the book. – Jason Feb 13 '11 at 18:37