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!
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!