Ε-δ definition of a limit

[chapsticks]

Homework Statement

use the elipson-delta of a limit to justify the answer of: show proof
lim x->0 1/(x+1)

Homework Equations

lim x->0 1/(x+1)=1

The Attempt at a Solution

lim x->0 1/(x+1)

I did some work but after that I don't know how to keep going.

if 0<|x-c|<δ => |f(x)-<ε|, then lim x->c f(x)=L
0<|x-0|<δ=> |1/(x+1)-1|
scratchwork: |1/(x+1)-1|<ε
|x/x+1|<ε
|x|/|x+1|<ε
|x|<|x+1|ε
restrict |x|<1/2
-1/2<x<1/2=>1/2<x+1<3/2=>1/2 <|x+1|
|x|<|x+1|ε
|x|<1/2 ε
δ=min{1/2,1/2ε}
how do I find the proof ??

 

[SammyS]

Homework Statement

use the elipson-delta of a limit to justify the answer of: show proof

lim x->0 1/(x+1)=1

The Attempt at a Solution

I did some work but after that I don't know how to keep going.

if 0<|x-c|<δ => |f(x)-L| < ε then lim x->c f(x)=L
0<|x-0|<δ=> |1/(x+1)-1|
scratchwork: |1/(x+1)-1|<ε
|x/x+1|<ε
|x|/|x+1|<ε
|x|<|x+1|ε
restrict |x|<1/2
-1/2<x<1/2=>1/2<x+1<3/2=>1/2 <|x+1|
|x|<|x+1|ε
|x|<1/2 ε
δ=min{1/2,1/2ε}
how do I find the proof ??

For the most part, reverse the steps in the scratch-work part.

Of course, start with

Let ε > 0​

Then define: δ=min{1/2,1/2ε}

δ ≤ 1/2 should lead back to 1/2 <|x+1| if 0 < |x-0| < δ . What does that say about 1/|x+1|

δ ≤ ε/2 and 0 < |x-0| < δ give |x| < ε/2

This along with your conclusion about 1/|x+1|, should pretty much do the trick.​

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