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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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    @mariano, this calculus class homework is submitted and processed by webassign.net. $T$he 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 f2011-02-13

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

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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.2011-02-13