Prove the limit????

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June 9th 2012, 03:47 PM
sluggerbroth

Prove the limit????

Prove the limit (as x goes to 3) x^(2)=9
June 9th 2012, 04:07 PM
Plato

Re: Prove the limit????

Quote:

Originally Posted by sluggerbroth

Prove the limit (as x goes to 3) x^(2)=9

You are kidding us are you not?
You are "pulling the proverbial leg" are you not?
If $x$ is close to $3$ but $x\ne 3$ , what is $x^2$ close to?
How does one prove that?
June 9th 2012, 05:45 PM
sluggerbroth

Re: Prove the limit????

I wish I were kidding but the epsilon delta proofs are painfully simple? please enlighten me plato!!!
June 9th 2012, 08:14 PM
Prove It

Re: Prove the limit????

I expect Plato was extremely surprised that you could not use the fact that $\displaystyle \begin{align*} \lim_{x \to a}f(x) = f(a) \end{align*}$ if the function is continuous at $\displaystyle \begin{align*} x = a \end{align*}$ . If you need an $\displaystyle \begin{align*} \epsilon - \delta \end{align*}$ proof, you need to STATE it.
June 10th 2012, 03:34 PM
HallsofIvy

Re: Prove the limit????

The problem is that we cannot read your mind! You said "Prove that $\lim_{x\to 3} x^2= 9$ " and there are many different ways to do that, depending on what you know. One method, if you know about continuous functions and, in particular, that all polynomials are continous for all x, then you have immediately that the limit is $3^2= 9$ , as Plato said. If that was not what you wanted, well, we have no way of knowing that unless you tell us exactly what you want. That is one reason why we repeatedly ask people to show what they have tried- so that we will at least know what basis they are starting from.

If you need to use an "epsilon-delta" proof then we needed to know that from the start. To have $|f(x)- L|= |x^2- 9|= |(x-3)(x+3)|< \epsilon$ . That can be reduced to $|x-3||x+3|< \epsilon$ or $|x-3|< \frac{\epsilon}{|x+3|}$ so to be able to say $|x-3|< \delta$ implies that, you need a lower bound on $\frac{\epsilon}{|x+3|}$ and so an upper bound on |x+ 3|.

We are looking at values of x close to 3 so we might as well look at |x- 3|< 1. That is,-1< x- 3< 1 so 5< x+ 3< 7.
June 10th 2012, 08:37 PM
Prove It

Re: Prove the limit????

Quote:

Originally Posted by HallsofIvy

One method, if you know about continuous functions and, in particular, that all polynomials are continous for all x, then you have immediately that the limit is $3^2= 9$ , as Plato said.

I must be a really good ventriloquist, and Plato a really good dummy ><