Why does $\lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = 1$? Why does $\displaystyle\lim_{n \to \infty}  \sqrt{\frac{n}{n+1}} = 1$? Shouldn't it be undetermined? 
 A: Note that for every $\epsilon >0$ we can find $N>0$ such that for all $n>N$, $$\left|\frac{n}{n+1}-1\right|<\epsilon$$
A: Recall what the definition of a limit means. Informally, beyond a large value of $n$, $\sqrt{\dfrac{n}{n+1}}$ should be very close to $1$. To put it formally, given an $\epsilon > 0$, you need to find $N(\epsilon)$ such that for all $n >N(\epsilon)$, we have
$$1- \sqrt{\dfrac{n}{n+1}} < \epsilon \tag{$\dagger$}$$
Now pick $N(\epsilon) = \left\lceil \dfrac1{\epsilon(2-\epsilon)} \right\rceil$ and show that for $n>N(\epsilon)$, $\dagger$ is satisfied.
A: $$ \lim_{x \rightarrow \infty} \sqrt{\frac{x}{x + 1}} ~ = ~  \lim_{x \rightarrow 0^+} \sqrt{\frac{1}{x + 1}} ~ = ~ 1$$
A: Alternative: $$ \lim_{n \to \infty} \sqrt {\dfrac {n}{n+1}} = \sqrt {\lim_{n \to \infty} \left( \dfrac {n}{n+1} \right)} = \sqrt {1} = 1. $$
A: This problem is quite simple if you know L'Hopitals rule, which states:
$\displaystyle\lim_{n\to\infty} \frac{f(x)}{g(x)}=\lim_{n\to\infty}\frac{f'(x)}{g'(x)}$
if $f(x)→∞$ and $g(x)→∞$ and both are continuous. 
$\displaystyle\lim_{n\to\infty} \sqrt\frac{n}{n+1}= \sqrt{\lim_{n\to\infty}\left(\frac{n}{n+1}\right)}=\sqrt{1}=1$
since (by L'Hopitals rule): 
$\displaystyle\lim_{n\to\infty}\frac{n}{n+1}=
\lim_{n\to\infty}\frac{\frac{d}{dn}n}{\frac{d}{dn}(n+1)}=
\lim_{n\to\infty}\frac{1}{1}=1$
A: It is very well determined.
$\lim_{n \to \infty}
\frac{n}{n+1}
= 1$
(since
$1-\frac{n}{n+1}
= \frac{1}{n+1}
\to 0$).
Since $\sqrt{}$ is continuous at $1$,
if $f(n) \to 1$
as $n \to \infty$,
$\sqrt{f(n)}
\to \sqrt{1}
= 1$.
