Given real $x\geq1$, show that: $\lim_{n\to\infty} (2\sqrt[n] x -1)^n=x^2.$ I proved it using this method. Is this proof correct? $
\lim_{n\to\infty} (2\sqrt[n] x -1)^n = 
\lim_{n\to 0+} e^{\frac{\log(2x^n -1)}{n}}
=\lim_{n\to 0+} e^{\frac{2x^n\log x}{2x^n-1}}
= e^{2\log x} = x^2.$
 A: The result is correct but you should clarify that step
$$\cdots \lim_{n\to 0+} e^{\frac{\log(2x^n -1)}{n}}
=\lim_{n\to 0+} e^{\frac{2x^n\log x}{2x^n-1}}=\cdots$$
As an alternative
$$(2\sqrt[n] x -1)^n =(1+(2\sqrt[n] x -2))^n=\left[(1+(2\sqrt[n] x -2))^{\frac1{2\sqrt[n] x -2}}\right]^{n(2\sqrt[n] x -2)}\to e^{\log x^2}=x^2$$
indeed
$$y=2\sqrt[n] x -2 \to 0 \implies (1+y)^\frac1y \to e$$
and
$$n(2\sqrt[n] x -2)=2\frac{ x^\frac1n -1}{\frac1n} \to2\log x$$
A: $$L=\lim_{n \rightarrow \infty} (2~ x^{1/n}- 1)^n= x \lim_{n \rightarrow \infty} \left( 2-\frac{1}{(1+y)^{1/n}}\right)^n, y>0, x=1+y. $$ Then
$$L=x \lim_{n \rightarrow \infty} \left( 2- (1-y/n)\right)^n=\lim_{n \rightarrow \infty} x (2-(1-ny/n)= x (1+y)= x^2.$$
Here we use binomial approximation twice: $\frac{1}{(1+y)^{1/n}}\approx 1-y/n$ and $(1-y/n)^{n} \approx 1-ny/n.$
A: $(2\sqrt[n] x -1)^n =(2 x^{\frac{1}{n}} -1)^n= (2 e^{\log(x) \frac{1}{n}} -1)^n \\=(2 \left(1 + \log(x) \frac{1}{n}+(\log(x) \frac{1}{n})^2+  ...\right) -1)^n\\= (1 +  \frac{1}{n}\left(  2\log(x) \right))^n\to e^{2\log(x) }= x^2 $
Here we have used the definition $\lim_{n\to\infty} \left( 1+ \frac{z}{n}\right)^n = e^z$
