<s>[INST] Calculating $\lim_{n\to\infty}\sqrt[n]{ \sqrt[n]{n} - 1 }$ I know that   
$$\lim_{n\to\infty} \sqrt[n]{ \sqrt[n]{n} - 1 } = 1,$$
but I'm unable to prove it. I could easily estimate that it's at most $1$, but my best estimation from below is that the limit is greater than $0$.
Doing this from the definition doesn't lead me anywhere either.
 [/INST] We can manipulate limits as long as we are sure that the part we are jumping over is bounded. For example:
$$\lim\limits_{x \to \infty} f(x)^{g(x)} = 
\lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty}g(x)}$$
as long as $f(x)$ is bounded.
First we attack $\lim\limits_{n \to \infty} n^{\frac{1}{n}}$. We make a substitution $x=\frac{1}{n}$
$$\lim\limits_{n \to \infty} n^{\frac{1}{n}}=\lim\limits_{x \to 0} (\frac{1}{x})^x=\lim\limits_{x \to 0} \frac{1}{x^x}$$
Now this one is elementary $\lim\limits_{x \to 0} x^x = 1$ so we have $\lim\limits_{n \to \infty} n^{\frac{1}{n}}=1$
We continue
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}(1-\frac{1}{\sqrt[n]{n}})}=\lim\limits_{n \to \infty}n^{\frac{1}{n^2}} (1-\frac{1}{n^{\frac{1}{n}}})^{\frac{1}{n}}$$
Since $n^{\frac{1}{n^2}}=(n^{\frac{1}{n}})^{\frac{1}{n}}$ we have as well $\lim\limits_{n \to \infty} n^{\frac{1}{n^2}}=1$
That leaves
$$\lim\limits_{n \to \infty} \sqrt[n]{\sqrt[n]{n}-1}=\lim\limits_{n \to \infty} (1-(\frac{1}{n})^{\frac{1}{n}})^{\frac{1}{n}}$$
The same substitution and we have 
$$\lim\limits_{x \to 0} (1-x^x)^{x}$$
$x^x$ has a minimum at $\frac{1}{e}$ and $1 > 1-x^x > 0$ for $0 < x < 1$ meaning it is bounded and $1-x^x$ is not reaching 0 for any $x$ in this region so we can examine
$$\lim\limits_{x \to 0} (1-x^x)^{x}=\lim\limits_{x \to 0} x^{x^2}(\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}$$
This is because we can write $x^{x^2}=(x^x)^x$
Obviously both limits are equal $\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (1-x^x)^{x}$.
Notice that
$$\lim\limits_{x \to 0} (\frac{1}{x^x}-1)^{x}=\lim\limits_{x \to 0} (x^{-x}-1)^{x}=\lim\limits_{x \to 0^{-}} (x^{x}-1)^{-x}=\lim\limits_{x \to 0^{-}} \frac{1}{(x^{x}-1)^{x}}$$
which is to say that the right limit is equal to reciprocal left (if it exists). This means that the unique limit, if it exists, is 1 since left and right limits are then the same.
But, $(1-x^x)^x$ is continuous at 0, $x^x$ is continuous and $a^x$ is continuous, meaning the limit is indeed 1.
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