# Proving limit by definition

#### goody

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

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#### lev888

##### Senior Member
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

##### New member
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

##### Elite Member
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 $$\displaystyle \delta = g(\epsilon)$$

#### HallsofIvy

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