Limit of n-th root of $1/n$ I'm struggling to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}$$
I have tried taking log,
$\log(\frac{1}{n})^{\frac{1}{n}}=\frac{1}{n}\log({\frac{1}{n}})$
and setting $t=\frac{1}{n}$ and rewite the desired limit as
$e^{\lim_{t\rightarrow 0}t\log(t)}$,
and I'm stuck here because $\log$ is not defined at $0$.
Did I miss something?
Thank you for reading.
 A: Just for the fun of it, here is another one:
Consider $\sum\limits_{n=1}^{\infty}\frac{1}{n}x^{n}$. This diverge at $x=1$ and converge at $x=-1$, hence the radius of convergence must be $1$. Using root test on this series give the radius of convergence to be $\frac{1}{\limsup\limits_{n\rightarrow\infty}(\frac{1}{n})^{\frac{1}{n}}}$ which show that $\limsup\limits_{n\rightarrow\infty}(\frac{1}{n})^{\frac{1}{n}}=1$.
Now consider $x^{x}=e^{x\ln(x)}$. Differentiate give $(1+\ln(x))e^{x\ln(x)}$ which is negative as long as $\ln(x)<-1$ which is $x<\frac{1}{e}$. Hence $(\frac{1}{n})^{\frac{1}{n}}$ is eventually monotone. That combine with $\limsup\limits_{n\rightarrow\infty}(\frac{1}{n})^{\frac{1}{n}}=1$ showed that $\lim\limits_{n\rightarrow\infty}(\frac{1}{n})^{\frac{1}{n}}=1$.
A: $\left(\dfrac{1}{n}\right)^{\frac{1}{n}} = \dfrac{1}{\sqrt[n]{n}} \to 1$ because:
$1 < \sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1...\cdot \sqrt{n}\cdot \sqrt{n}} < \dfrac{1+1+...+1 + 2\sqrt{n}}{n} = \dfrac{n-2+2\sqrt{n}}{n} = 1 -\dfrac{2}{n} + \dfrac{2}{\sqrt{n}}$. Apply squeeze theorem the answer follows.
A: By the L'Hôpital's rule we have
$$\lim_{t\to0}t\ln t=\lim_{t\to0}\frac{\ln t}{\frac1t}=\lim_{t\to0}\frac{\frac1t}{-\frac1{t^2}}=-\lim_{t\to0}t=0$$
A: this is a standard matter.  I hope you've worked out a conjecture first by using a computer programme or just plugging in some numbers on the formula. (You should, as it is a healthy thing).
Either you continue as you started (continuous variable $t$), and write your limit as
$$\frac{log(t)}{1/t},$$ (i.e. a quotient of infinities) and apply whichever result you please to your newly-found indeterminate limit (L'Hopital), or you can use a standard result on sequences.
I suggest e.g. D'Alembert's Criterion or the Quotient Criterion (look up Apostol's Calculus, for instance, its chapter on sequences). (Stoltz's theorem also works).
A nicer way, which can accompany you for the rest of your life, is as follows.
Consider $1+x_n=n^{1/n},$ and write the identity
$$n=(1+x_n)^n.$$
You can use Newton's binomial theorem here, and establish (choosing your favourite term, I'll choose the degree-2 term) that, for instance,
$$n> \frac{n(n-1)}{2} x_n^2 \geq 0.$$
After isolating, you'll see immediately what the expected result is.
Your sequence is very much related to this one.
There are many resources available online, e.g.
http://planetmath.org/proofoflimitofnthrootofn
REGARDING POLYNOMIALS AND EXPONENTIALS:  Actually, there are recipes such as this one: the exponential always beats a polynomial, or a power of $x$, as $x\to +\infty.$  The argument would be very similar.  Say, you want to show that, for $a>1$,
$$\frac{a^x}{x^p}$$ tends to infinity; you may restrict to $x=n$, and then compare (write $a=1+b$, for $b>0$)
$$(1+b)^n= 1 + nb + \cdots + \frac{n(n-1)\ldots (n-p+1)}{(p+1)!} b^{p+1} + \cdots,$$ which clearly is bigger than $n^p$, for $n \to \infty$.
I leave it as an exercise for you to see what happens as $x\to +\infty$ (take the integer part $n=n(x)$ of $x$, and bound numerator and denominator appropriately comparing $f(x)$ to $f(n)$).
