Simple delta proof for calc 1 class help

[Drman]

Anyone want to help me with a very simple proof? This is only for calculus 1 and we just went over epsilon delta proof of a limit(non infinity). I'm getting stuck trying to solve one though. Limit as x-> 3 (2/x+3) =1/3 Should be straight forward but not sure how to format the proof as correctly as possible. Thanks for any help, would appreciate it solved if possible.

 

[HallsofIvy]

You say "help" you with the proof and say you are not sure of the "formatting". Okay, what have you done that we could help you with? Show what you have done and I am sure many people will point out where you are right or wrong and where you might word something better.

You do know that "\(\displaystyle \lim_{x\to a} f(x)= L\) if and only if, given \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) such that if \(\displaystyle |x- a|< \delta\) then \(\displaystyle |f(x)- L|< \epsilon\)", right? So apply that to this case where a= 3, f(x)= 2/(x+ 3) (I believe that is what you mean, not "2/x+ 3"), and L= 1/3.

What does \(\displaystyle |f(x)- L|<\epsilon\) using your f(x) and L?

 

[Drman]

Yes sorry the part where I'm stuck in just my scratch work to make the proof work is

|-(x-3)/3(x+3)|< Epsilon or my |f(x)-L|< epsilon I want to make it equal to | x- a|<delta or |x-3|< delta how do I make them equivalent in a proof?


Sorry about that

 

[HallsofIvy]

That's pretty much just algebra. \(\displaystyle \frac{2}{x+ 3}- \frac{1}{3}= \frac{6}{3(x+ 3)}- \frac{x+ 3}{3(x+ 3)}= \frac{3- x}{3(x+ 3)}\). The absolute value of that gives \(\displaystyle \left|\frac{3- x}{3(x+ 3)}\right|=\)\(\displaystyle \left|\frac{1}{3(x+ 3)}\right||x- 3|\). You want to find, for any \(\displaystyle \epsilon> 0\), to find \(\displaystyle \delta> 0\) such that "if \(\displaystyle |x- 3|< \delta\) then \(\displaystyle \left|\frac{1}{3(x+ 3)}\right||x- 3|< \epsilon\). What you need to do is find bounds for that \(\displaystyle \left|\frac{1}{x(x+ 3)}\right|\). Since we are talking about the "limit as x goes to 3" we can start by assuming that x is close to 3: say 2< x< 4 just because those numbers are easy. Then 5< x+ 3< 7 and (5)(2)= 10< x(x+ 3)< 4(7)= 28. From that \(\displaystyle \frac{1}{28}< \frac{1}{x(x+ 3)}< \frac{1}{10}\). Those numbers are all positive so we don't need to worry about the absolute value. Now we know that, as long as x is between 2 and 4, that is, as long as x- 3 is between -1 and 1 so |x- 3|< 1, we have
\(\displaystyle \frac{1}{28}|x- 3|< \left|\frac{1}{3(x+3)}\right||x- 3|< \frac{1}{10}|x- 3|\).

 

[Drman]

That's pretty much just algebra. \(\displaystyle \frac{2}{x+ 3}- \frac{1}{3}= \frac{6}{3(x+ 3)}- \frac{x+ 3}{3(x+ 3)}= \frac{3- x}{3(x+ 3)}\). The absolute value of that gives \(\displaystyle \left|\frac{3- x}{3(x+ 3)}\right|=\)\(\displaystyle \left|\frac{1}{3(x+ 3)}\right||x- 3|\). You want to find, for any \(\displaystyle \epsilon> 0\), to find \(\displaystyle \delta> 0\) such that "if \(\displaystyle |x- 3|< \delta\) then \(\displaystyle \left|\frac{1}{3(x+ 3)}\right||x- 3|< \epsilon\). What you need to do is find bounds for that \(\displaystyle \left|\frac{1}{x(x+ 3)}\right|\). Since we are talking about the "limit as x goes to 3" we can start by assuming that x is close to 3: say 2< x< 4 just because those numbers are easy. Then 5< x+ 3< 7 and (5)(2)= 10< x(x+ 3)< 4(7)= 28. From that \(\displaystyle \frac{1}{28}< \frac{1}{x(x+ 3)}< \frac{1}{10}\). Those numbers are all positive so we don't need to worry about the absolute value. Now we know that, as long as x is between 2 and 4, that is, as long as x- 3 is between -1 and 1 so |x- 3|< 1, we have
\(\displaystyle \frac{1}{28}|x- 3|< \left|\frac{1}{3(x+3)}\right||x- 3|< \frac{1}{10}|x- 3|\).

Ah Thanks so much for the help this make sense