hello, this is not really homework but i am more of a pioneer of knowledge as you can say, i know it sounds corny, but i am having trouble on this calculus problem:
lim x -> 3 (-x^2)/((x^2)- 6x + 9)
when using direct substitution i get -9/0
when factoring i get the same result
therefore is it infinity?
if it is not infinity how would i come about finding my answer
One has to be careful to call anything dividing by zero as infinity.
Firstly, infinity is not a number. And while it is true that on a real line, infinity is not the same as 'minus infinity' in complex analysis, all infinity is considered as one whether you go North to hit infinity, or go South until infinity or East until infinity. This one point compactification (or stenographic projection) is not hard to visualize when "infinity" is the point on the other side of "the globe".
So you might want to call lim x -> 3 (-x^2)/((x^2)- 6x + 9) as negative infinity. But remember "infinity" is not a number, so if you have
lim x -> 3 (-9)/(x-3)
then you might have a trouble because this may seem to be -9/0, but it could be either infinity or negative infinity. Because on the graph of y=(-9)/(x-3), the line x=3 is a vertical asymptote and as x approaches 3 from the right, the limit is negative infinity but when as approaches 3 from the left the limit is positive infinity. so we have to consider left and right limit. But because they are so different, it is better to say lim x -> 3 (-9)/(x-3) does not exist.
Again "limit does not exist" can come in different categories too. If you are taking an analysis course or some rigorous Calculus 1 course, you will learn the different kind of "limit does not exist".
Hope that helps.