prove that $\lim\limits_{n\to \infty} 2^{\frac{1}{\sqrt{n}}} =1 $ without using continuity of functions Prove that $$ \lim_{n\to\infty} 2^{\frac{1}{\sqrt{n}}} =1 $$ I tryed to solve this using the sandwich theorem and the fact that $\lim 2^{\frac{1}{{n}}} =1 $. We did not learn continuity of functions so I am not alowed to use this. Any suggestions?
First: $2^{1/n} \to 1.$ Proof: Let $2^{1/n} = 1 + a_n.$ Then $a_n &gt; 0$ and the binomial theorem gives

$$2 = (1+a_n)^n = 1 + na_n + \cdots \ge 1 + na_n.$$

Thus $0&lt; na_n \le 1 \implies a_n \to 0 \implies 2^{1/n}\to 1.$

As for $2^{1/\sqrt n}:$ Let $\epsilon&gt;0.$ From the above there exists $m$ such that $1&lt; 2^{1/m} = 2^{1/\sqrt {m^2}}&lt;1+\epsilon.$ Because $p&lt; q \implies 2^p&lt; 2^q,$ we have $1&lt; 2^{1/\sqrt {n}}&lt;1+\epsilon$ for $n\ge m^2.$ Thus $2^{1/\sqrt n}\to 1.$
Think about this intuitively, as n approaches infinity, what does the value of 1/sqrt(n) become? Well, I think we agree that as n approaches infinity, the square root of n becomes infinity. So, what is 1 divided a really really big number? It is a really small number aka zero; 2 to the zero-th power is one. Therefore, the limit is 1.