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

1. ## 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

2. ## Re: elipson-delta definition of a limit to justify the answer Please help not sure ab

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 $\displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon$, then $\displaystyle \lim_{x \to c}f(x) = L$.

In this case, you want to show $\displaystyle \lim_{x \to 0}\frac{1}{x + 1} = 1$, so you need to show that $\displaystyle 0 < |x - 0| < \delta \implies \left|\frac{1}{x + 1} - 1\right| < \epsilon$.

Some scratch-work...

\displaystyle \begin{align*} \left|\frac{1}{x + 1} - 1 \right| &< \epsilon \\ \left|\frac{x}{x + 1}\right| &< \epsilon \\ \frac{|x|}{|x + 1|} &< \epsilon \\ |x| &< |x + 1|\epsilon \end{align*}

Now suppose we restrict $\displaystyle |x| < \frac{1}{2}$ (say), then $\displaystyle -\frac{1}{2} < x < \frac{1}{2} \implies \frac{1}{2} < x + 1 < \frac{3}{2} \implies \frac{1}{2} < |x + 1|$, therefore

\displaystyle \begin{align*} |x| &< |x + 1|\epsilon \\ |x| &< \frac{1}{2}\epsilon \end{align*}

So if we let $\displaystyle \delta = \min\left\{\frac{1}{2}, \frac{1}{2}\epsilon\right\}$ and reverse the process, you will have your proof.

3. ## Re: elipson-delta definition of a limit to justify the answer Please help not sure ab

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

4. ## Re: elipson-delta definition of a limit to justify the answer

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