Proving limit by definition

[goody]

Hi, can anybody help me with this two limits? I have to prove them by the definition of limit. Thanks.

 

[lev888]

Please post your favorite definition of a limit. Do you understand it? Have you worked through an example of a proof the question requires?

 

[goody]

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.

 

[Steven G]

Given $$\epsilon>0, \ \exists \ \delta >0$$ such that $$\left| f(x) - a\right|<\epsilon \ whenever \ 0<\left| x - a\right|<\delta$$
Now do whatever you can to turn $$\left| f(x) - a\right|<\epsilon \ into \ \left| x - a\right|< g(\epsilon).$$
Finally define $$\delta = g(\epsilon)$$

 

[HallsofIvy]

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 $\displaystyle |(x^2+1)-2|= |x^2- 1|= |(x-1)(x+1)|< |3\left(\frac{\epsilon}{3}\right)|= \epsilon$.