Real Analysis: Proving a fuction is continuous

[SNOOTCHIEBOOCHEE]

I Apologize in advance for the amount of questions i plan on asking you guys this year. Real Analysis is my first upper division math class and i have not trained my mind to think abstractly enough yet.

Homework Statement

i) Show that f(x) = x^3 is continuous on R by using (Epsilon)−(Delta) property

The Attempt at a Solution

Ok so there is basically one part that i get stuck at, but here is what i have so far


Suppose that limit Xn= Xo

So we have lim f(Xn)=lim[Xn^3]= Xo^3 =f(Xo)
hence f is continuous for each Xo in R

|f(X)-f(Xo)|=|X^3 - Xo^3|= |(X - Xo)(X^2 + XXo + Xo^2)| <$$\epsilon$$
|X-Xo|<$$\delta$$


Now here is where i am stuck. How exactly do i find epsilon? Have I even done this proof correctly so far? Also after I find epsilon, what exactly do i do...

Please Reply Over...

Mike

 

[HallsofIvy]

So you have $|x- x_0||x^2+ x_0x+ x_0^2|< \epsilon$ and the problem is that second absolute value.

Choose some convenient distance from x₀. For example, let's suppose that
$|x- x_0|< 1$. That is the same as saying $x_0-1< x< x_0+ 1$. Use that to find an upper bound on $x^2+ x_0x+ x_0^2$, say M, so you have
$|x-x_0||x^2+x_0x+ x_0^2|< M|x-x_0|$. That will be less than $\epsilon$ as long as $M|x-x_0|< \epsilon$ which says $|x-x_0|< \frac{\epsilon}{M}$. All you have to do is decide how large M has to be.

 

[SNOOTCHIEBOOCHEE]

So you have $|x- x_0||x^2+ x_0x+ x_0^2|< \epsilon$ and the problem is that second absolute value.

Choose some convenient distance from x₀. For example, let's suppose that
$|x- x_0|< 1$. That is the same as saying $x_0-1< x< x_0+ 1$. Use that to find an upper bound on $x^2+ x_0x+ x_0^2$, say M, so you have
$|x-x_0||x^2+x_0x+ x_0^2|< M|x-x_0|$.

Ok the highlighted statement is giving me some trouble... wouldn't M just be equal to $|x^2 + x_0x+ x_0^2|$ ?

 

[HallsofIvy]

No, because M has to be a number, not a function of x. If x is between $x_0-1$ and $x_0+ 1$ what is the largest possible value of $x^2+ x_0x+ x_0^2$?