Basic limit problem, requesting backup

[M4z3]

It says: Use the formal definition of limit to verify the indicated limit.

lim x->1 1/x+1 = 1/2

in this sheet i got with answers, i dont understand the steps they're explaining so i would apreciate
some kind of a newbie explanation since im not that good at absolute values.

Regards

 

[HallsofIvy]

(What you wrote, 1/x+1, would, by the usual "precedence of operations", be interpreted as (1/x)+ 1 which does NOT have limit 1/2. You mean 1/(x+1).)

The "formal definition" of "the limit, as x goes to a, of f(x)" is this:
\(\displaystyle \lim_{x\to a} f(x)= L\) if and only if, given any \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) so that if \(\displaystyle |x- a|< \delta\) then \(\displaystyle |f(x)- L|< \epsilon\).

So start from \(\displaystyle |f(x)- L|= |\frac{1}{x+ 1}- \frac{1}{2}\)\(\displaystyle = |\frac{2}{2(x+1)}- \frac{x+1}{2(x+1)}|= |\frac{1- x}{2(x+ 1)}|= |1- x|\frac{1}{|2(x+1)|}< \epsilon\).

The |1- x|= |x- 1| is what you need and you can write \(\displaystyle |1- x|< |2(x+ 1)|\epsilon\) so we need a bound on |2(x+ 1)|. We can say that if |x- 1|< 1, then -1< x- 1< 1 or 0< x< 2. 0< 2x< 4, 2< 2x+ 2< 6. So we can say that \(\displaystyle 2\epsilon< |2x+2|\epsilon\).

Now, if you choose \(\displaystyle \delta= 2\epsilon\) and \(\displaystyle |x- 1|< \delta= 2\epsilon\), can you reverse that to get \(\displaystyle |\frac{1}{x+1}- \frac{1}{2}|< \epsilon\)?

 

[M4z3]

(What you wrote, 1/x+1, would, by the usual "precedence of operations", be interpreted as (1/x)+ 1 which does NOT have limit 1/2. You mean 1/(x+1).)

The "formal definition" of "the limit, as x goes to a, of f(x)" is this:
\(\displaystyle \lim_{x\to a} f(x)= L\) if and only if, given any \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) so that if \(\displaystyle |x- a|< \delta\) then \(\displaystyle |f(x)- L|< \epsilon\).

So start from \(\displaystyle |f(x)- L|= |\frac{1}{x+ 1}- \frac{1}{2}\)\(\displaystyle = |\frac{2}{2(x+1)}- \frac{x+1}{2(x+1)}|= |\frac{1- x}{2(x+ 1)}|= |1- x|\frac{1}{|2(x+1)|}< \epsilon\).

The |1- x|= |x- 1| is what you need and you can write \(\displaystyle |1- x|< |2(x+ 1)|\epsilon\) so we need a bound on |2(x+ 1)|. We can say that if |x- 1|< 1, then -1< x- 1< 1 or 0< x< 2. 0< 2x< 4, 2< 2x+ 2< 6. So we can say that \(\displaystyle 2\epsilon< |2x+2|\epsilon\).

Now, if you choose \(\displaystyle \delta= 2\epsilon\) and \(\displaystyle |x- 1|< \delta= 2\epsilon\), can you reverse that to get \(\displaystyle |\frac{1}{x+1}- \frac{1}{2}|< \epsilon\)?

Thanks alot, you totally crushed it, i got it all took a while, but i suck at math ^_^. Much obliged