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.
$\left(\dfrac{1}{n}\right)^{\frac{1}{n}} = \dfrac{1}{\sqrt[n]{n}} \to 1$ because:

$1 &lt; \sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1...\cdot \sqrt{n}\cdot \sqrt{n}} &lt; \dfrac{1+1+...+1 + 2\sqrt{n}}{n} = \dfrac{n-2+2\sqrt{n}}{n} = 1 -\dfrac{2}{n} + \dfrac{2}{\sqrt{n}}$. Apply squeeze theorem the answer follows.
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,
$$n&gt; \frac{n(n-1)}{2} x_n^2 \geq 0.$$

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 $a&gt;1$,

$$\frac{a^x}{x^p}$$ tends to infinity; you may restrict to $x=n$, and then compare (write $a=1+b$, for $b&gt;0$)

$$(1+b)^n= 1 + nb + \cdots + \frac{n(n-1)\ldots (n-p+1)}{(p+1)!} b^{p+1} + \cdots,$$ 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 $f(n)$).