# understanding definition of limit for x_n= [(n+1)/n]^0.5; where x_n is the n-th term in the sequence.

$x_n= \sqrt{\frac{n+1}{n}}$where $x_n$ is the $n^{th}$ term in the sequence.

Now, $x_n \rightarrow 1$ as $n\rightarrow\infty$.

Then applying definition of limit: $$\left|\sqrt{\frac{n+1}{n}}- 1 \right|= \frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}<\frac{1}{2n}<\epsilon$$

whenever $n>N=1/(2\epsilon)$.

-The difficulty for me is inability to understand how $1/(2n)$ was deduced.

Thank you for assistance and apologies for poor notations

the equation in the code should read after absolute value equation = {[(n+1)/n]-1}/{([(n+1)/n]^0.5)+1}<1/2n< epsilon

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I've edited your question, adding LaTeX code to make it more readable. please accept the edit :) –  Yoni Rozenshein Sep 26 '12 at 8:40
thank you! I've been trying to figure out how the code works. Any guides out there you may link me to? –  student101 Sep 26 '12 at 8:43
The code is called LaTeX. I guess you can search for LaTeX tutorials but the way I learned it is by just starting to write stuff in it, and whenever I didn't know how to do something, I searched for how to do that thing until I remembered. –  Yoni Rozenshein Sep 26 '12 at 8:47
@student101: Is the recent edit correct? I don't think so. –  Babak S. Sep 26 '12 at 8:48
ill rewrite what i wrote initially, not sure what happened to the edits! –  student101 Sep 26 '12 at 8:49

Multiply the expression $\left(1+\frac{1}{n}\right)^{1/2}-1$ by $1$, in the form $\displaystyle\frac{\left(1+\frac{1}{n}\right)^{1/2}+1}{\left(1+\frac{1}{n}\right)^{1/2}+1}$.

The numerator is then $\left(1+\dfrac{1}{n}\right)-1$, which simplifies to $\dfrac{1}{n}$.

The denominator $\left(1+\dfrac{1}{n}\right)^{1/2}+1$ is $\gt 2$, since $1+\dfrac{1}{n}\gt 1$, and therefore its square root is $\gt 1$.

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I see that since $n(n+1)>n$ then $\sqrt{n(n+1)}+n>2n$ and so $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$$ But, $$\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|$$ Therefore $$\bigg|\sqrt{\frac{n+1}{n}}-1\bigg|=\bigg|\frac{\frac{n+1}{n}-1}{\sqrt{\frac{n+1}{n}}+1}\bigg|=\bigg|\frac{1}{\sqrt{n(n+1)}+n}\bigg|<\frac{1}{2n}$$

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I understand now that 1/2n makes sense in the inequality...but im wondering, why choose exactly 1/2n to be the term on the RHS... is the purpose of this to make the RHS a function of only 1 'n' term? –  student101 Sep 26 '12 at 12:02
@student101: Mostly right. In fact you need a logical and proper relation, containing $n$, just to make a connection between $n$, $\epsilon$ from one side and then with $N$. So, you get $\epsilon=1/2n$ and then $n=1/2\epsilon$. This is what you took as $N$. Of course $1/2n$ might be something different for example $\epsilon=1/3n$. –  Babak S. Sep 26 '12 at 13:47