Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\lim_{n\to \infty}\frac{n}{n+1} = 1$

Prove using epsilon delta.

share|improve this question
    
Use $\displaystyle \frac{n}{n+1} = 1 - \frac{1}{n+1}$ and use the Archimedean property. –  Andrew Salmon May 27 '13 at 17:22
    
What is your definition of "epsilon-delta" when you approach infinity? The $0 <|x - c| < \delta$ part doesn't seem to work. –  Henry Swanson May 27 '13 at 17:23
    
we had a quiz on this question today, our TA told us to assume that lim(x->infinity) 1/n = 0 since not everyone has learned the Archimedean property (me being one of them). and using the assumption, derive an epsilon-delta proof. –  Nick Gong May 27 '13 at 17:26
3  
You want to show that for all $\epsilon$ there exists N such that if $n>N$, $|\frac{n}{n+1}-1|=|\frac{1}{n+1}|<\epsilon$ What does this tell you you should pick for N (in terms of $\epsilon$)? –  Zen May 27 '13 at 17:31

1 Answer 1

up vote 4 down vote accepted

By your notation I believe you're talking about the sequence $(a_n)$ of elements of $\mathbb{R}$ defined by:

$$a_n =\frac{n}{n+1}$$

Now, limit for sequences has the following definition: "given $\varepsilon >0$ there's some $n_0 \in \mathbb{N}$ such that if $n > n_0$ then $|a_n - L|<\varepsilon$". So we want some $n_0$ such that:

$$\left|\frac{n}{n+1} - 1\right|<\varepsilon$$

Rewrite this as:

$$\left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right|$$

But $n$ is natural so that the thing inside of the module sign is already positive and so we want in truth that:

$$\frac{1}{n+1}<\varepsilon \Longrightarrow n>\frac{1-\varepsilon}{\varepsilon}$$

This part is the deduction part. Now we prove, we say: given $\varepsilon > 0$ take $n_0 =(1-\varepsilon)/\varepsilon$, then we have that for $n > n_0$:

$$\left|\frac{n}{n+1}-1\right| = \left|\frac{1}{n+1}\right| = \frac{1}{n+1}$$

But $n>n_0$ so that $1/(n+1) < 1/(n_0 + 1)$ and hence:

$$\left|\frac{n}{n+1} - 1\right| < \frac{1}{n_0 + 1} = \frac{1}{\frac{1-\varepsilon}{\varepsilon} + 1} = \varepsilon$$

The important part for you to note is that first we deduce which $n_0$ works, after that we throw this part away usually and just say: "take this $n_0$" so that we show that it really works as predicted.

share|improve this answer
    
Very nicely done, +1! –  dreamer May 27 '13 at 17:41
    
Thank you soo much! really appreciate the help –  Nick Gong May 28 '13 at 16:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.