Another proof of limit by definition

[ILikePizza]

Homework Statement

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.

Homework Equations

The definition of a limit.

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!

 

[HallsofIvy]

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)

 

[ILikePizza]

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)

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

 

[HallsofIvy]

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&lt; \epsilon, you will want to use (x-c)(x^2+ cx+ c^2&lt; (x-1)(3c^2- c+ 1)&lt; \epsilon. Of course, in order to guarantee |x-c|< 1 you will have to say "let \delta be the smaller of 1 or ...