why is it that x=1 is not considered a vertical asymptote for a particular function even though when plugged into that function it gives the form 0/0. I'm confused since that's an indeterminate form, so why is it not a vertical asymptote?

Question

anonymous · Answer

because if you get $\frac{0}{0}$ when you substitute, that means you can factor and cancel

anonymous · Answer

for example in 
$$\frac{x^2-9}{x-3}$$  $x=3$ is not a vertical asymptote because this is the same as 
$$\frac{(x+3)(x-3)}{x-3}=x+3$$

tkhunny · Answer

No, that's not quite it.  It means x = 1 is NOT IN THE DOMAIN.

$\dfrac{x^{2}-9}{x-3} \ne x+3$ for x = 3

tkhunny · Answer

"plugged into" - Please get this concept out of your head.  It has no place in mathematics.

For starters, you MAY NOT substitute the value x = 1 into your expression if it makes the denominator zero (0).  If it makes then denominator zero (0), it's not in the Domain.

Second, Around asymptotes, graphs exhibit asymptotic behavior.  There is nothing about your graph around x = 1 that can be described as asymptotic.  It just snuggles up to it and stops.  That's not an asymptote.

anonymous · Answer

see that's what I thought when you cancel out, it automatically becomes an asymptote since your value would not be defined otherwise. the given question was find the vertical asymptotes of f(x) = $$\frac{ x^{3}-1 }{ x^{3}-x } $$ and when you simplify, (x-1) 'cancels' out and the factors on the bottom are x and (x+1) and the exam answer keys states that only x=0 and x=-1 are asymptotes

campbell_st · Answer

the situation you describe can be called a point of discontinuity.... the curve doesn't exist at that point... 

as example is 

$$y = \frac{x^2 -2x + 1}{x^2 + 4x - 5} $$

(x -1) is a common factor,  but the point is x =1 would normally be a vertical asymptote. 

so it is just a point where the curve doesn't exist

campbell_st · Answer

a point of discontinuity can be graphed as

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