<s>[INST] 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
 [/INST] 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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