<s>[INST] Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem with m and x being whole numbers, for x greater or equal to 1 and m greater or equal to 0. My work so far:
We define $f(x) := x^\frac{1}{m+1}$. From the MVT it follows that there exists an $x_0$ between a and b (with b > a) such that: $$\frac{f(b)-f(a)}{b-a} =  f'(x_0)$$  
From choosing $b = x$ and $a = x - 1$ it follows that:
$$x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1} = \frac{1}{(m+1)x_0^\frac{m}{m+1}}$$  
Multiplication with $x^\frac{m}{m+1}$ results in:  
$$x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{x^\frac{m}{m+1}}{(m+1)x_0^\frac{m}{m+1}}$$  
I intuitively want to take the limit to infinity of the left-hand side, but I can't justify why the $x$ and $x_0$ terms will cancel-out. I'd also to be happy to hear of any methods that don't use the MVT.
 [/INST] When $x\to \infty$ since $x-1<x_0<x$ we also have that $x_0 \to \infty$ and
$$x-1<x_0<x \iff \frac{x-1}x<\frac{x_0}x<1$$
then by squeeze theorem $\frac{x_0}x\to 1$, thus
$$\frac{x^\frac{m}{m+1}}{(m+1)x_0^\frac{m}{m+1}}=\frac1{m+1} \left(\frac{x}{x_0}\right)^{\frac{m}{m+1}} \to\frac1{m+1}$$
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