# Thread: Prove the limit using the epsilon delta method?

1. ## Prove the limit using the epsilon delta method?

I am kinda of new to the methods of proving limits with the epsilon delta method, and I can figure out simple limits, but more general ones seem to be quite difficult. Heres the question at hand:

Prove, by epsilon-delta methods, the limit of the following:

$\lim_{x \to y} \frac{x^n -y^n}{x-y}$

I know the rules for minipulating limits, as in "the limit of a sum of functions is the sum of the limits of those functions" and "the limit of the product of two functions is the product of the limits of those functions", ect. I've gone at this problem and haven't had much luck, I've realized that you can factor out a $(x-y)$ from the top, leaving the second factor of the expanded factorization of $x^n - y^n$ next to it. But, I'm not sure if I should cancel the $(x-y)$'s because I feel like they will be crucial and usefull in implying $|\frac{x^n - y^n}{x-y} - L|< \epsilon$ from $0 < |x-y| < \delta$. Any help would be much appriciated, and also if anybody can give a general method of how to tackle general limits like this one, I would appreciate that also.

2. HINT:

x^n - y^n=(x-y){x^(n-1)+x^(n-2)y+...+y^(n-1)}

3. Originally Posted by Also sprach Zarathustra
HINT:

(x-y)^n=(x-y){x^(n-1)+x^(n-2)y+...+y^(n-1)}
Didn't you mean something like

$x^n-y^n=(x-y)\left\{x^{n-1}+x^{n-2}y+...+y^{n-1}\right\}$

4. Originally Posted by Failure
Didn't you mean something like

$x^n-y^n=(x-y)\left\{x^{n-1}+x^{n-2}y+...+y^{n-1}\right\}$

I get that much, I just dont't know how to quantify the $x^{n-1}+x^{n-2}y+...+y^{n-1}$ in terms of an inequality, which would alow me to properly define $\delta$ in a manner which implies the epsilon inequality. Say I make the stipulation that $|x-y|<1$, this leads to:

$y-1

I'm trying to get the middle expression (x in the above inequality) to be the expaned factor of the $x^n-y^n$ (the longer factor, not the $(x-y)$ factor) Because then I could find a number $m$, probably in terms of x and y I'm not sure, which is greater then the longer factor for all x and y, thus giving me an inequality of:

$|x-y|(m)>f(x)$

something like the above, I may have made some mistake, something seems off, but I you can get my point and what I mean. Something that would allow me to define $\delta$ properly, any more hints?

5. First guess the L...

L=n*x^(n-1)

now look at:

|{x^n -y^n}/{x-y} - n*x^(n-1)|=.... {use the fact that |x-y|<delta}

6. Originally Posted by Also sprach Zarathustra
First guess the L...

L=n*x^(n-1)

now look at:

|{x^n -y^n}/{x-y} - n*x^(n-1)|=.... {use the fact that |x-y|<delta}
Okay, I see what your getting at, and I also see how the next steps of the proof would be played out. But what about this; at some other point two days ago I remember that I came across a thought about limits; and that thought was the following assertion:

If $f(x)$ is a polynomial function, that is non rational, then the following always holds:

$\lim_{x \to a} f(x) = f(a)$

I took some time to try and prove this (that is assuming this guess is actually true) and I think I succesfully proved it. I'm not sure if this assertion is true, which obviously implies that I am not confident in the validity of my proof. Therefore, if you see any flaws, or know this assertion is false, please let me know. Anyway, my proof started by using the fact (this fact I proved in the previous chapter of the text book I'm working from) using the fact that any polynomial function $f(x)$ can be written in the form:

$f(x) = (x-a)g(x)+b$

Given this fact, if I have not made any mistakes, I beleive it follows that:

$0 < |x-a| < \delta \;, \;\;\; and \;\;\; | f(x) - f(a) | < \epsilon$

if I require that:

$\delta = \min(1, \; \frac{\epsilon}{g(x)})$

or something of the sort. Do the above steps prove that for a polynomial function $f(x)$:
$\lim_{x \to a} f(x) = f(a)$ ??

And if so, adding this together with the facts that if:

$\lim_{x \to a} f(x) = l$ and $\lim_{x \to a} g(x) = m$

then:

$\lim_{x \to a}[ f(x) + g(x) ] = l + m \;\;\; , \;\; \lim_{x \to a}[ f(x)g(x) ] = (l)(m)$ and if:

$m \neq 0$ then:

$\lim_{x \to a} \frac{1}{g(x)} = \frac{1}{m}$

Then am I correct in assuming that the original limit I was attempting to obtain from the function, after canceling the $(x^n-y^n)$'s, the resluting function:

$f(x) = (x^{n-1} + x^{n-2}y + x^{n-3}y + \cdot \; \cdot \; \cdot \; + xy^{n-3} + xy^{n-2} + y^{n-1})$

is simply given as:

$\lim_{x \to a} f(x) = f(a)$

?????????