# Another proof of limit by definition

1. Oct 26, 2008

### ILikePizza

1. The problem statement, all variables and given/known data
Prove that as x --> c, lim (x^2) = (c^2) using only the definition. What does this tell you about x --> c, lim (x^3) = (c^3)? x --> c, lim (x^4) = (c^4) ? Prove it.

2. Relevant equations
The definition of a limit.

3. The attempt at a solution
Let $\epsilon > 0$ be given and let $\delta=\sqrt{c^2 + \epsilon} - c$. Then, if $0 < |x - c| < \delta$, then
$|x^2 - c^2| = |x - c||x + c| < \delta|x + c|$.
Since $|x +c| --> 2c$, and $|\delta| --> 0$ their product can be made as small as you want, and we are done.

I think c^4 case goes right off c^2, but I have no idea where to start with c^3. Also, is there a way to make the proof above a little more rigorous?

Thanks!

2. Oct 26, 2008

### HallsofIvy

Staff Emeritus
Not "\$". Use [ tex ] and [ /tex ] (without the spaces which I put in so you could see the code itself.

x3- c3= (x- c)(x2+ cx+ c2)

3. Oct 26, 2008

### ILikePizza

So x-c = $$\delta$$, and x2+ cx+ c2 --> 3c^2.

So if $$\delta$$ = $$\frac{\epsilon}{3c^2}$$, that's good. But how do I write this rigorously?

Thanks again

4. Oct 27, 2008

### HallsofIvy

Staff Emeritus
Since you were specifically asked to do this "using only the definition", saying x2+ cx+ c2--> 3c2 may not be allowed.

What you can do is this: Assume that |x-c|< 1 so c-1< x< c+1. Then, assumin,temporarily, that 0<c-1, c2- 2c-1< x2< c2+ 2c+ 1 and c2- c< cx< c2+ c. Adding those, 3c2-c+ 1< x2+cx+ c2< 3c2+ 3c+ 1. Since you want $(x-c)(x^2+ cx+ c^2< \epsilon$, you will want to use $(x-c)(x^2+ cx+ c^2< (x-1)(3c^2- c+ 1)< \epsilon$. Of course, in order to guarentee |x-c|< 1 you will have to say "let $\delta$ be the smaller of 1 or ...