2
$\begingroup$

Obviously, $\displaystyle \lim_{n\to\infty} \sqrt{1 + {1 \over n}} = 1$ since $$\lim_{n\to\infty}{1 \over n} = 0, \tag{$*$}$$ but how might I show that $\displaystyle \lim_{n\to\infty} \sqrt{1 + {1 \over n}} = 1$ using basic properties of limits? My first thought was to say $$L = \lim_{n\to\infty}\sqrt{1 + {1 \over n}} \implies L^2 = \lim_{n\to\infty}\left(1 + {1 \over n}\right) = 1 \implies L = \pm 1.$$

But obviously we get the extraneous solution of $L = -1$, which is not true since $\sqrt{n} > 0$ for all $n\in\mathbb N$.

This is actually a smaller step in a much larger limit problem I'm attempting. I can take $(*)$ as an absolute fact since it is a result we've proved in my textbook already. However, there is no mention of the limit of a square root.

$\endgroup$
  • 4
    $\begingroup$ $\lim\limits_{n\to\infty}\sqrt{1+\frac1n}=\sqrt{\lim\limits_{n\to\infty}1+\frac1n}$ $\endgroup$ – barak manos Sep 30 '16 at 9:26
4
$\begingroup$

Hint: $$1\le \sqrt{1+\frac1n}\le 1+\frac1n$$ This solution uses the squeeze theorem, which may (or may not) be thought as an "evaluation using limit laws" as you requested.

$\endgroup$
  • 1
    $\begingroup$ This hint gives me just the information I need to build an argument from. I appreciate it! $\endgroup$ – Decaf-Math Sep 30 '16 at 9:54
4
$\begingroup$

Since $$ |\sqrt{1+n^{-1}}-1|=\biggl|\frac{(\sqrt{1+n^{-1}}-1)(\sqrt{1+n^{-1}}+1)}{\sqrt{1+n^{-1}}+1}\biggr|=\biggl|\frac{n^{-1}}{\sqrt{1+n^{-1}}+1}\biggr|\le \frac12 n^{-1}, $$ we have that $\sqrt{1+n^{-1}}\to1$ as $n\to\infty$.

$\endgroup$
  • 1
    $\begingroup$ This answer (as well as the others currently posted) is perfectly acceptable since it proves that the limit goes to $1$ as $n\to\infty$. However, it isn't as satisfactory as I'd hoped since we don't really evaluate it using limit laws. It's technically a class on analysis, so answers like this aren't above what I know and can use, but the unit we are in is about using limit laws to deduce the answer rather than proving it since we already proved some of the limit laws. I suppose I should have asked the linked answer with $a=1$ for clarity. $\endgroup$ – Decaf-Math Sep 30 '16 at 9:42
  • 1
    $\begingroup$ @pyrazolam I think the most natural way is to use the continuity of the mapping $x\mapsto \sqrt x$ defined for $x\ge0$. Then you could simply show that $\lim_{n\to\infty}\sqrt{1+n^{-1}}=\sqrt{\lim_{n\to\infty}(1+n^{-1})}=1$. barak manos wrote this in the comment. I am not sure if there exists a way to evaluate this limit using only the basic properties of limits. $\endgroup$ – Cm7F7Bb Sep 30 '16 at 9:53
4
$\begingroup$

By definition, $\lim a_n=1$ means that: $$ \forall \epsilon>0, \exists N>0: \forall n>N, \ \ 1-\epsilon\leq a_n\leq 1+\epsilon $$ or equivalently: $$ \forall \epsilon>0, \exists N>0: \forall n>N, \ \ |a_n -1|\leq \epsilon $$ Now, for your sequence $a_n=\sqrt{1 + {1 \over n}}$, for any given $\epsilon>0$ it suffices to pick any integer $N>\frac{1}{\epsilon(\epsilon+2)}$ to plug in the above definition and see it works.

P.S.: Notice that, $\forall \epsilon>0: \ \ $ $1-\epsilon\leq 1<\sqrt{1 + {1 \over n}}\leq 1+\epsilon\Leftrightarrow n\geq\frac{1}{\epsilon(\epsilon+2)}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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