Delta/Epilson proof

Hello,

After reading the thread's sticky on how to do DE proofs, I attempted one of my home work questions so could someone let me know if I did it right? I think I did but my confidence in my skills at this stage is low so I feel I need lots of reassurance

Prove lim x^2-4x-12=9
x->-3

in my rough work I basically get it to factor to x^2-4x-21 which is |x-7||x-(-3)|
delta equals epilson/|x-7| and I label |x-7| as M since Delta cannot be defined in terms of x
x is -4<x<-2 if |x-(-3)|<1
M is therefore -2
and Delta becomes E/-2

I won't write out the full proof but basically at the end I say my delta is min {1, E/-2} and that 0<|x-(-3)| <Delta implies that |x^2-4x-21|<Epsilon

Thanks!

Quote:

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Originally Posted by funnybabe

Hello,

After reading the thread's sticky on how to do DE proofs, I attempted one of my home work questions so could someone let me know if I did it right? I think I did but my confidence in my skills at this stage is low so I feel I need lots of reassurance

Prove lim x^2-4x-12=9
x->-3

in my rough work I basically get it to factor to x^2-4x-21 which is |x-7||x-(-3)|
delta equals epilson/|x-7| and I label |x-7| as M since Delta cannot be defined in terms of x
x is -4<x<-2 if |x-(-3)|<1

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You are almost correct to this point.
Build your other factor: $\displaystyle -11<x-7<-9$.
From which you see that $\displaystyle |x-7|<11$.
So let $\displaystyle \delta = \min \left\{ {1,\frac{\varepsilon }{{11}}} \right\}$.

so when you do the other factor -11<x-7<-9, how come you don't compute that in absolute value? I am assuming you are subbing in x=-4 and then when x=-2. if x=-2, the absolute value is 9 and wouldn't that be the minimum?
btw, where do you get to use the delta and epsilon characters?
thanks!

my other question is basically isn't the minimum -2? it sounds like you are saying the minimum is -4. just a little confused. then Delta=E/9?

Quote:

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Originally Posted by funnybabe

so when you do the other factor -11<x-7<-9, how come you don't compute that in absolute value? I am assuming you are subbing in x=-4 and then when x=-2. if x=-2, the absolute value is 9 and wouldn't that be the minimum?
btw, where do you get to use the delta and epsilon characters?
thanks!

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If $\displaystyle -7<a<-3$ then we know that absolutely $\displaystyle |a|<7~.$ Do we not?

If $\displaystyle -4<a<6$ then we know that $\displaystyle |a|<6~.$

If $\displaystyle -4<a<2$ then we know that $\displaystyle |a|<4~.$

So if $\displaystyle -11<x-7<-9$ then we know that $\displaystyle |x-7|<11~.$

We are using the maximum bound.

I think I get it. Basically just because -4 is lower then -2, the absolute value of -4 is greater then the absolute value of -2 so go with the maximum value of the absolute. is this correct?

Quote:

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Originally Posted by funnybabe

I think I get it. Basically just because -4 is lower then -2, the absolute value of -4 is greater then the absolute value of -2 so go with the maximum value of the absolute. is this correct?

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Given that $\displaystyle a<x<b$ then $\displaystyle |x|<\max\{|a|,|b|\}$.