Proof $\lim_{x\rightarrow\infty}\left(2\sqrt[x]a-1\right)^x=a^2$ 
Proof that for $a\ge1$ we get
$\lim_{x\rightarrow\infty}\left(2\sqrt[x]a-1\right)^x=a^2$

$\lim_{x\rightarrow\infty}\left(2\sqrt[x]a-1\right)^x=\lim_{x\rightarrow\infty}\left(\frac{4\sqrt[x]{a^2}-1}{2\sqrt[x]a+1}\right)^x$
I know that $\lim_{x\rightarrow\infty}\sqrt[x]a=1$
but I don't know how I can use it, power $x$ bothers a lot
 A: Setting $x=\frac 1t$ and considering $t\to 0^+$ you get
$$\left(2\sqrt[x]a-1\right)^x
\stackrel{x=\frac 1t}{=} (2a^t-1)^{\frac 1t}
$$
Now, take the logarithm and use L'Hospital for example:
\begin{eqnarray} \frac{\log(2a^t-1)}{t}
& \stackrel{L'Hosp.}{\sim} & \frac{2\log a\cdot a^t}{2a^t-1} \\
& \stackrel{t\to 0^+}{\longrightarrow} & \log a^2
\end{eqnarray}
Hence, $\lim_{x\rightarrow\infty}\left(2\sqrt[x]a-1\right)^x= e^{\log a^2} = a^2$.
A: How about working this way. 
\begin{eqnarray}
\mathcal L &=&\lim_{x \to +\infty} \left(2\sqrt[x]{a}-1\right)^x=\\
&=&\lim_{x\to +\infty}e^{x\log \left(2{a}^{1/x}-1\right)}=\\
&=&\lim_{x\to +\infty}e^{x\log(1+2a^{1/x}-2)}=\\
&=&\lim_{x\to +\infty}e^{2\frac{a^{1/x}-1}{1/x}}=\\
&=&\lim_{x\to +\infty}e^{2\log a}=a^2.
\end{eqnarray}
Where I used
$$\frac{\log(1+\alpha (x))}{\alpha(x)} \to 1$$
and
$$\frac{a^{\alpha(x)}-1}{\alpha(x)}\to \log a,$$
when $\alpha(x) \to 0$.

Alternatively, as you propose in comment
\begin{eqnarray}
\mathcal L &=& \lim_{x\to +\infty}[1+(2a^{1/x}-2)]^x=\\
&=&\lim_{x\to +\infty}\left\{\underbrace{\left[1+(2a^{1/x}-2)\right]^{\frac{1}{2a^{1/x}-2}}}_{\to e}\right\}^{x(2a^{1/x}-2)}=\\
&=&\lim_{x\to +\infty}e^{2\frac{a^{1/x}-1}{1/x}}=a^2
\end{eqnarray}
