Epsilon - Delta Proof Question

lim ((1/(x+4)) = 1/2

x-> -2

or

a=-2

L=1/2

f(x)= (1/(x+4))

The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.

This is the part that is confusing me because I can solve any other limit question.

The options for answers are:

a)[0,min{1,4epsilon})

b)(0,min{1,6epsilon}]

c)[0,min{1,2epsilon})

d)(0,min{1,2epsilon}]

e)(0,min{1,4epsilon})

Re: Epsilon - Delta Proof Question

Quote:

Originally Posted by

**poipoipoi10** lim ((1/(x+4)) = 1/2

x-> -2

The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.

This is the part that is confusing me because I can solve any other limit question.

The options for answers are:

a)[0,min{1,4epsilon})

b)(0,min{1,6epsilon}]

c)[0,min{1,2epsilon})

d)(0,min{1,2epsilon}]

e)(0,min{1,4epsilon})

Notice that .

If what can you say about

Re: Epsilon - Delta Proof Question

Sigh... I'm confused -.-

I guess that

-1<|x+2|<1

-3<x<-1

1<x+4<3

sooooo uh yeah... haha

Re: Epsilon - Delta Proof Question

Quote:

Originally Posted by

**poipoipoi10** Sigh... I'm confused -.-

Here a hint.

Re: Epsilon - Delta Proof Question

So would the answer be D???

Because if you plug 1 as the |x+4|, from the equation you gave me: 1|x+2|/2|x+4|<epsilon

I would get |x+2|<2*epsilon

I don't really know if i've done that correctly, I've been doing work all day and my head is swimming with different subjects :(

Re: Epsilon - Delta Proof Question

Quote:

Originally Posted by

**poipoipoi10** lim ((1/(x+4)) = 1/2

x-> -2

or

a=-2

L=1/2

f(x)= (1/(x+4))

The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.

This is the part that is confusing me because I can solve any other limit question.

The options for answers are:

a)[0,min{1,4epsilon})

b)(0,min{1,6epsilon}]

c)[0,min{1,2epsilon})

d)(0,min{1,2epsilon}]

e)(0,min{1,4epsilon})

To show you need to prove . So you need to show .

Scratch work:

But we can't let because is only supposed to be a function of , not a function of . But if we remember that the act of taking a limit means that we are showing that moving closer to the value of in both directions moves closer to the value of in both directions, that means we can choose a value of as close to as we like, then make it move closer. So say we want to start off with no more than unit away from , then . Therefore

So we can let and reverse each step to get your proof.