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Why does $\displaystyle\lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = 1$? Shouldn't it be undetermined?

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  • $\begingroup$ Hint: Bring the limit inside , L'H rule. $\endgroup$
    – Amzoti
    Nov 20 '13 at 2:32
  • $\begingroup$ Ah okay, you are allowed to do that though? Even if it was inside ln ? $\endgroup$
    – Joshua Ree
    Nov 20 '13 at 2:42
  • $\begingroup$ You have to follow the rules of the power law and continuity in order for it to be legal. $\endgroup$
    – Amzoti
    Nov 20 '13 at 2:48
  • $\begingroup$ Why did you think it should be undetermined? $\endgroup$ Nov 20 '13 at 3:13
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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 $n \to \infty$, $\sqrt{f(n)} \to \sqrt{1} = 1$.

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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.

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Note that for every $\epsilon >0$ we can find $N>0$ such that for all $n>N$, $$\left|\frac{n}{n+1}-1\right|<\epsilon$$

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  • $\begingroup$ And $\sqrt x$ is continuous at $x=1$. $\endgroup$
    – Pedro Tamaroff
    Nov 20 '13 at 3:35
  • $\begingroup$ @PedroTamaroff: Thanks for the time. I thought (s)he knew this point and was able to manage the points all together to find the answer as others did. $\endgroup$
    – Mikasa
    Nov 20 '13 at 6:45
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$$ \lim_{x \rightarrow \infty} \sqrt{\frac{x}{x + 1}} ~ = ~ \lim_{x \rightarrow 0^+} \sqrt{\frac{1}{x + 1}} ~ = ~ 1$$

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  • $\begingroup$ Cleanest answer here in my humble opinion. $\endgroup$
    – Squirtle
    Nov 20 '13 at 4:50
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Alternative: $$ \lim_{n \to \infty} \sqrt {\dfrac {n}{n+1}} = \sqrt {\lim_{n \to \infty} \left( \dfrac {n}{n+1} \right)} = \sqrt {1} = 1. $$

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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):

$\displaystyle\lim_{n\to\infty}\frac{n}{n+1}= \lim_{n\to\infty}\frac{\frac{d}{dn}n}{\frac{d}{dn}(n+1)}= \lim_{n\to\infty}\frac{1}{1}=1$

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    $\begingroup$ Using L'Hopitals rule here is a kind of sybaritism )) $\endgroup$
    – GusRustam
    Nov 20 '13 at 13:16

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