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Hi;
I have the rational function:
This factors into:
The problem says you cannot simplify this at x= -3 because the denominator is 0, so you cannot cancel out the (x+3)
Taking the limit:
I think because the limit exists the cancellation is fine
Now at the other pole x=1/3 the limit is
So maybe at x=1/3 the cancellation is not possible. What do you think?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Like you said, you can cancel it everywhere except x=-3.
The function is undefined at x=-3, but it is defined everywhere close to it, so the limit of -1/10 exists.
The cancellation is allowed at x=1/3, but doing that gets you 1/0, so it's still undefined.
This time there also isn't a limit, because the limits from each direction don't agree.
Why did the vector cross the road?
It wanted to be normal.
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Hi maths;
I'm not following you here, if the limit equals -1/10 at x = -3 doesn't that mean that the function is defined at x= -3, so shouldn't I be able to do the cancellation. Sorry, I am drawing a blank here.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Just because the limit of f(x) as x approaches c exists does not mean that f(c) is equal to this limit.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Since there is a limit at x=3, you can safely cancel as long as you keep in mind that the function is undefined at that point. More formally, you can rewrite your function as
It doesn't matter that your function is undefined at x = 1/3, since that's the case both before and after the cancellation. You didn't change that fact by simplifying the function.
Wrap it in bacon
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Thanks mathsy, Ricky and TheDude. It is a clearer now.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym
From the beggining, the substitution of x=1/3 in the rational function is 1/0 so the limit is either undefined (∞) or does not exist. In our example the left hand side limit is (-∞) since the values of are <0 , and the right hand side limit is (∞) since the values of are >0 , so the limit does not exist. Also the cancellation does not work since there is not in the numerator that can be cancelled with (3x-1) which exists in the denominator.
Best Regards
Riad Zaidan
Hi Riad Zaidan;
Thanks, your right about that one too.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Dear bobbym
thanks for your attention , and I am very pleased with you.
Best Wishes
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