How can I find $\displaystyle \lim_{n\to\infty} \sqrt{1 + {1 \over n}}$ using basic properties of limits?

Question

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.

Answer 1

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.

Answer 2

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 $n\to\infty$.

Answer 3

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: \ \ $ $1-\epsilon\leq 1<\sqrt{1 + {1 \over n}}\leq 1+\epsilon\Leftrightarrow n\geq\frac{1}{\epsilon(\epsilon+2)}$