show that the limit of $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$ alright so im supposed to show that the limit of $\displaystyle \lim_{n \to \infty} n^{\frac{1}{n}} = 1$
, ive alwasy had problems with math questions of this type: "show" and never really knew what to do, but since its $n^{\frac{1}{n}}$ thats basically $n$th-root of $n$ when $\lim \to \infty =1$ of that function so yeah :)
I have no idea how to go forward with this problem, but I managed to figure out that if $n=2$ then I get $1.4$, if $n=5$ then I get $1.38$ and if $n=1000$ then I get $1.007$ so it seems like im getting closer and closer to $1$ as $x$ approach larger numbers?? ok anyway! thanks for tips and help :D
 A: Hint: Note that $n^{1/n} \geq 1$ for $n\geq 1$ so there is a sequence of non-negative reals $\delta(n)$ s.t. $n^{1/n} = 1 + \delta (n)$ for each $n$. Show that $\delta(n) \to 0$ as $n\to \infty$ by raising both sides to the power of $n$.

Solution: In particular, note that:
$n = (1+\delta (n))^n =  1+n\delta (n) + \dfrac{n(n-1)}{2}[\delta(n)]^2 + … + [\delta (n)]^n$
$\implies n \geq \dfrac{n(n-1)}{2}[\delta(n)]^2$ (As all other terms are non-negative)
$\implies 0\leq \delta (n) \leq \sqrt{\dfrac{2}{n-1}} \to 0$ as $n\to \infty$
$\implies \delta (n) \to 0$ as $n\to \infty$ (By squeezing)
$\implies 1+ \delta (n) \to 1$ as $n \to \infty$
Hence $\displaystyle\lim_{n\to \infty} n^{1/n} = 1$, as desired.
A: By AM-GM and Sandwich Theorem
$\frac{1 + 1 + 1 + \dots + \sqrt{n} + \sqrt{n}}{n} \geq \sqrt[n]{n} \geq 1$
$\lim \limits_{n \to \infty} \frac{n-2 + 2 \sqrt{n}}{n}$
$\lim \limits_{n \to \infty} 1 - \frac2n + \frac{2}{\sqrt{n}}$
$\therefore \lim \limits_{n \to \infty} \sqrt[n]{n} = 1$
A: Hint: take Log and use L'hospitale rule.
A: $$\mbox{To calculate: }\lim_{n\to\infty}n^{1/n}$$
Let $x=n^{1/n}$. Then, $\ln(x)=\dfrac{\ln(n)}{n}$. So,
$$\lim_{n\to\infty}n^{1/n}=\exp\left(\lim_{n\to\infty}\dfrac{\ln(n)}{n}\right)\\
\exp\left(\lim_{n\to\infty}\dfrac{\ln(n)}{n}\right)=\exp\left(\lim_{n\to\infty}\dfrac{1}{n}\right)\mbox{(by L'Hopital's rule.) This you can show goes to $1$.}$$
A: Here is another way and one of the simplest. 
Since $\liminf (n+1)/n\le \liminf n^{1/n}\le\limsup n^{1/n}\le \limsup(n+1)/n$. Then  $1\le \liminf n^{1/n}\le\limsup n^{1/n}\le 1$. 
Thus $\liminf n^{1/n}=\limsup n^{1/n}=1$ and so $\lim n^{1/n}=1$. 

It's worth to know that for  $(a_n) \subset \mathbb{R}^{>0}$, If $\lim a_{n+1}/a_n= L$ where $0\le L < \infty$, then $\lim a_n^{1/n}=L$. This last assertion is obvious from the fact that 
$$\liminf a_{n+1}/a_n \le \liminf a_n^{1/n}\le \limsup a_n^{1/n}\le \limsup {a_{n+1}/a_n}$$
