Prove Lim of x^2 as x approaches 3 = 9 with Epsilon/ Delta Definition

[TitoSmooth]

Prove Lim x^2=9. With the epsilon/delta definition of a limit.
x->3

My work so far. For every ε>0 there is a δ>0 such that

if 0<|x-3|<δ , Then |x^2-9|<ε

so, |(x-3)(x+3)|<ε

|x-3|* |x+3|∠ε

what do I do from here? My book is not very clear (Stewart Calculus 7ed) on how to do episolon/delta tolerance proofs on not so nice functions.

please help me.

 

[TitoSmooth]

and i also know that |x-3| is smaller then delta. I think this is a clue.

 

[wakefield]

Try writing [itex]|x-3||x+3|<\epsilon [/itex] as [itex]|x-3|<\frac{\epsilon}{|x+3|} [/itex] and then substituting that for [itex]\delta [/itex].

 

[TitoSmooth]

Try writing [itex]|x-3||x+3|<\epsilon [/itex] as [itex]|x-3|<\frac{\epsilon}{|x+3|} [/itex] and then substituting that for [itex]\delta [/itex].

so when sub everything back in I should get back the original? Therefore that last step was the answer? Or is there something more to it?

 

[Mark44]

Try writing [itex]|x-3||x+3|<\epsilon [/itex] as [itex]|x-3|<\frac{\epsilon}{|x+3|} [/itex] and then substituting that for [itex]\delta [/itex].

No, that's not how it works. If x is close to -3, then |x + 3| is close to zero, and you can't divide by zero. On the other hand, if x is close to 3, then |x + 3| is close to 6.

You have to make some assumptions on ##\delta## so that x is not too far away from 3.

 

[wakefield]

No, that's not how it works. If x is close to -3, then |x + 3| is close to zero, and you can't divide by zero. On the other hand, if x is close to 3, then |x + 3| is close to 6.

You have to make some assumptions on ##\delta## so that x is not too far away from 3.

Sure it is.

Spoiler

Say that [itex]|x-3|<1[/itex], then [itex]5<x+3<7[/itex]. Then we know that [tex]|x-3|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{7}.[/tex]Now we have a case where [itex]|x-3|<1[/itex] and a case where [itex]|x-3|<\frac{\epsilon}{7}[/itex]. So let [itex]\delta=min\{1,\frac{\epsilon}{7}\}[/itex].

It's not like a direct substitution but it's the right idea.

 

[TitoSmooth]

Sure it is.

Spoiler

Say that [itex]|x-3|<1[/itex], then [itex]5<x+3<7[/itex]. Then we know that [tex]|x-3|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{7}.[/tex]Now we have a case where [itex]|x-3|<1[/itex] and a case where [itex]|x-3|<\frac{\epsilon}{7}[/itex]. So let [itex]\delta=min\{1,\frac{\epsilon}{7}\}[/itex].

It's not like a direct substitution but it's the right idea.

how do we create the assumptions? Can I have an explanation in layman terms lol. This looks like the example in my book.

how did we restrict the values?

 

[Mark44]

Sure it is.

Spoiler

Say that [itex]|x-3|<1[/itex], then [itex]5<x+3<7[/itex]. Then we know that [tex]|x-3|<\frac{\epsilon}{|x+3|}<\frac{\epsilon}{7}.[/tex]Now we have a case where [itex]|x-3|<1[/itex] and a case where [itex]|x-3|<\frac{\epsilon}{7}[/itex]. So let [itex]\delta=min\{1,\frac{\epsilon}{7}\}[/itex].

It's not like a direct substitution but it's the right idea.

This is basically what I was talking about.