Thread: elipson-delta definition of a limit to justify the answer.

So Far I found the limit:

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

then this is the scratchwork:
1/(x+1)-1<E when 0< |x-0|< delta

Not sure if I did this inequality right:

|1/(x+1)-1|<E <==>|x/(x+1)|<E <==> |x/(x+1)|< 0E??

Do I have to find a number M??
1/|x+1| less than or equal to M

Proof:
|1/(x+1)-1|=|x|/|x+1| then I don't know how to keep going, not sure if it's 0E is correct

Originally Posted by LAPOSH42

So Far I found the limit:

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

then this is the scratchwork:
1/(x+1)-1<E when 0< |x-0|< delta

Not sure if I did this inequality right:

|1/(x+1)-1|<E <==>|x/(x+1)|<E <==> |x/(x+1)|< 0E??

Do I have to find a number M??
1/|x+1| less than or equal to M

Proof:
|1/(x+1)-1|=|x|/|x+1| then I don't know how to keep going, not sure if it's 0E is correct

If , then .

In this case, you want to show , so you need to show that .

Some scratch-work...



Now suppose we restrict (say), then , therefore



So if we let and reverse the process, you will have your proof.

If you want to have more examples of proves using definition then I recommand you to take a look here:
http://www.mathhelpforum.com/math-he...ofs-47767.html

Proof: let E>0 and define delta= min {1/2,1/2E}
then if 0<|x-0|<delta

|1/(x+1)-1| = |x/x+1|
=|x|/|x+1|
<(1/2)?|x-0|
<(1/2)?2E
=E