Proving limit by definition
- Thread starter goody
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Yes, I know how definition of limit works and I practised a couple of examples which were pretty simple, but this is something new for me when x approaches -1. We have online school now so not everything is easy. I'd like to see how it works so I can practise another, maybe even more difficult examples.
Jomo
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Given \(\displaystyle \epsilon>0, \ \exists \ \delta >0\) such that \(\displaystyle \left| f(x) - a\right|<\epsilon \ whenever \ 0<\left| x - a\right|<\delta\)
Now do whatever you can to turn \(\displaystyle \left| f(x) - a\right|<\epsilon \ into \ \left| x - a\right|< g(\epsilon).\)
Finally define \(\displaystyle \delta = g(\epsilon)\)
Now do whatever you can to turn \(\displaystyle \left| f(x) - a\right|<\epsilon \ into \ \left| x - a\right|< g(\epsilon).\)
Finally define \(\displaystyle \delta = g(\epsilon)\)
HallsofIvy
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We want to show that \(\displaystyle \lim_{x\to -1} x^2+ 1= 2\). I would start by noting that \(\displaystyle |(x^2+ 1)- 2|= |x^2- 1|= |(x- 1)(x+ 1)|\).
Now, if |x- 1|< 1 (chosen pretty much arbitrarily) then -1< x- 1< 1 so 0< x< 2 and 1< x+ 1< 3. So |(x-1)(x+ 1)|< 3|x- 1|. That will be less than any given \(\displaystyle \epsilon> 0\), \(\displaystyle 3|x- 1|< \epsilon\) if \(\displaystyle |x- 1< \frac{\epsilon}{3}\).
Now, I notice that when Lev888 said"Please post your favorite definition of a limit. Do you understand it?" you responded that you understood it but did NOT post it!
The one I would use is "\(\displaystyle lim_{x\to a} f(x)= L\) if and only if, for any \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) such that if \(\displaystyle |x-a|<\delta\) then \(\displaystyle |f(x)- L|< \epsilon\)."
Here we can say that for any \(\displaystyle \epsilon> 0\) we can take \(\displaystyle \delta= \frac{\epsilon}{3}\). Then, if |x- 1| is less than or equal to the smaller of \(\displaystyle \delta= \frac{\epsilon}{3}\) and 1, so that both \(\displaystyle x- 1< \frac{\epsilon}{3}\) and \(\displaystyle x+1< 3\) then \(\displaystyle |(x^2+1)-2|= |x^2- 1|= |(x-1)(x+1)|< |3\left(\frac{\epsilon}{3}\right)|= \epsilon\).
Now, if |x- 1|< 1 (chosen pretty much arbitrarily) then -1< x- 1< 1 so 0< x< 2 and 1< x+ 1< 3. So |(x-1)(x+ 1)|< 3|x- 1|. That will be less than any given \(\displaystyle \epsilon> 0\), \(\displaystyle 3|x- 1|< \epsilon\) if \(\displaystyle |x- 1< \frac{\epsilon}{3}\).
Now, I notice that when Lev888 said"Please post your favorite definition of a limit. Do you understand it?" you responded that you understood it but did NOT post it!
The one I would use is "\(\displaystyle lim_{x\to a} f(x)= L\) if and only if, for any \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) such that if \(\displaystyle |x-a|<\delta\) then \(\displaystyle |f(x)- L|< \epsilon\)."
Here we can say that for any \(\displaystyle \epsilon> 0\) we can take \(\displaystyle \delta= \frac{\epsilon}{3}\). Then, if |x- 1| is less than or equal to the smaller of \(\displaystyle \delta= \frac{\epsilon}{3}\) and 1, so that both \(\displaystyle x- 1< \frac{\epsilon}{3}\) and \(\displaystyle x+1< 3\) then \(\displaystyle |(x^2+1)-2|= |x^2- 1|= |(x-1)(x+1)|< |3\left(\frac{\epsilon}{3}\right)|= \epsilon\).