Uniform convergence of $\; x^{\frac{1}{n}-1}\;\ \forall x\ge 1$ let $f_n(x)=x^{\frac{1}{n}-1}\;\ \forall x\ge 1$ and let $f(x)=x^{-1}\;\ \forall x\ge 1$, then:
$$\:\ (f_n)_{n\in \Bbb N}\;\ \text{converges uniformly to}\; f$$

$$\text{Let}\;\ \epsilon >0, \text{lets take}\;\ N=N_{\epsilon} (\text{figuring out...}),\; \text{then for}\;\ n>N
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\Rightarrow\;\ \big|x^{\frac{1}{n}-1}-x^{-1}\big|=x^{\frac{1}{n}-1}-x^{-1}\\
\text{because}\;\ x\ge1\;\; \text{and}\; -1\le \frac{1}{n}-1\; \forall n\in \Bbb N\ \Rightarrow x^{-1}\le x^{\frac{1}{n}-1}\;\ \forall n\in \Bbb N\;\ \forall x\ge 1$$
by the arquemedean property it follows that $\; n>x\;\ \forall x\ge 1\; \Leftrightarrow\ -x^{-1}<-\frac{1}{n}\;\ \forall x\ge 1$, so
$$\big|x^{\frac{1}{n}-1}-x^{-1}\big|=x^{\frac{1}{n}-1}-x^{-1}<x^{\frac{1}{n}-1}-\frac{1}{n}$$
and here got stuck since the only thing I can do is make $x^{\frac{1}{n}-1}\le 1$. Any ideas will be appreciated.
 A: One approach to show the uniform convergence is to write for $x\ge 1$
$$\left|\frac{x^{1/n}-1}{x}\right|=\frac{x^{1/n}-1}{x}<\epsilon \tag 1$$
The inequality in $(1)$ is equivalent to 
$$\frac{x}{(1+x\epsilon)^n}<1 \tag 2$$
Now, note that from Bernoulli's Inequality we have
$$\frac{x}{(1+x\epsilon)^n}<\frac{x}{1+nx\epsilon} \tag 3$$
If the right-hand side of $(3)$ is less than $1$, than certainly $(2)$ is satisfied.  Then, 
$$\frac{x}{1+nx\epsilon}<1\implies n>\frac{x-1}{x}\frac{1}{\epsilon}$$
Note that $(x-1)/x<1$.  Therefore, if we let $N=1/\epsilon$, then whenever $n>N$ we have 
$$\left|\frac{x^{1/n}-1}{x}\right|<\epsilon$$
And we are done!
A: $$\sup_{x\geq 1}\left|x^{\frac{1}{n}-1}-\frac{1}{x} \right|=\sup_{x\geq 1}\frac{1}{x}\left|x^{\frac{1}{n}}-1\right|\leq\frac{1}{\left(1+\frac{1}{n-1}\right)^n}\left(\frac{1}{n-1}\right)\underset{n\to\infty }{\longrightarrow } 0 $$
and thus the convergence is uniform. 
NB: The result looks a little bit magic, but it's not. I just set $$f_n(x)=x^{\frac{1}{n}-1}-\frac{1}{x}$$ and compute it's maximum.
