- #1
hmm?
- 19
- 0
Hello,
I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks.
Prove that lim x->-2 (x^2-1)=3
0<|f(x)-L|<epsilon whenever 0<|x-a|<delta
0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta
= |x^2-4|<epsilon
= |(x-2)(x+2)|<epsilon
= |x-2||x+2|<epsilon
= If |x-2|<C (C=constant) then |x-2||x+2|<C|x+2|
= C|x+2|<epsilon = |x+2|<epsilon/C
Assume |x+2|<1 so -1<x+2<1 = -3<x<-1 = -5<x-2<-3
After this step -5<x-2<-3, I'm inclined to set |x-2|<-3, but book states that |x-2|<5. I'm lost at this step, should it be|-5|? Anyways, any explanations would be great.
Thanks,
Chris
I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks.
Prove that lim x->-2 (x^2-1)=3
0<|f(x)-L|<epsilon whenever 0<|x-a|<delta
0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta
= |x^2-4|<epsilon
= |(x-2)(x+2)|<epsilon
= |x-2||x+2|<epsilon
= If |x-2|<C (C=constant) then |x-2||x+2|<C|x+2|
= C|x+2|<epsilon = |x+2|<epsilon/C
Assume |x+2|<1 so -1<x+2<1 = -3<x<-1 = -5<x-2<-3
After this step -5<x-2<-3, I'm inclined to set |x-2|<-3, but book states that |x-2|<5. I'm lost at this step, should it be|-5|? Anyways, any explanations would be great.
Thanks,
Chris