I recommend you read this for detailed explanation and working.
http://www.mathhelpforum.com/math-he...ta-proofs.html
You'll find example 3 to be especially helpful.
Hello , I have been having trouble with understanding what to do with those ε-δ Limits that involve an extra "x" factor in the result, I'll just write out the problem and perhaps you could correct or instruct me on the logic behind it.
Here, I'm kind of confused. I've seen the best explanation so far here: Created with Camtasia Studio 5 & I'd like to follow the method to get everything in working order.
1: Because of the |x - 2|< δ, (i.e. the x-->2 in the original limit), we know that x is very close to 2, just a bit bigger than it. Because this is so the |x + 3| term is very close to 5, just a bit above it. In fact, it will definitely be less than 6.
We can say with confidence that;
|x + 3||x - 2| < 6|x - 2|
2: Here I'm a bit unsure, can we say that;
6|x - 2| < ε
|x - 2|< ε/6 ? I think this works because although the gap from. say, 5.0000001 to being less than 6 (from the inequality) is big, when we divide ε by 6 we are creating more space.
3: Is there a way to make it look pretty? Is the answer that as long as δ = ε/6 we can guarantee the limit?
I recommend you read this for detailed explanation and working.
http://www.mathhelpforum.com/math-he...ta-proofs.html
You'll find example 3 to be especially helpful.
I already had that pdf on my laptop & I never understood it until I followed the video I linked to but I may understand it now.
If I take my example;
What I can do is look back at |x - 2| & if I want to bound it within 1 unit it would be equivalent to bound the |x + 3| factor in the denoinator within a distance of 1 as well;
Then, because the limit is going near 2 we'll take the |x + 3| and realise it's near 5, but because of the constriction in |x - 2| of being within 1 unit we'll do the same for the |x + 3|
Then, if we take 6 as the min. i.e. δ = (1,ε/6)
So;
I haven't made any careless mistakes or assumptions have I?