Proving $\lim_{n \to \infty} \left(2x^{\frac{1}{n}}-1\right)^{n}=x^2$

Question

Prove that $$\lim_{n \to \infty} \left(2x^{\frac{1}{n}}-1\right)^{n}=x^2$$ $\forall$ $x \ge 1$

My try:

For $x=1$ The proof is Trivial

If $x \gt 1$ we have by Binomial Theorem

$$\left(2x^{\frac{1}{n}}-1\right)^{n}=2^nx-\frac{nx2^{n-1}}{x^{\frac{1}{n}}}+\frac{\binom{n}{2}x2^{n-2}}{x^{\frac{2}{n}}}-\cdots$$

Any way here?

Answer 1

We have that $$\begin{align}\left(2x^{\frac{1}{n}}-1\right)^{n}&=\left(2e^{\frac{\ln(x)}{n}}-1\right)^{n}= \left(2\left(1+\frac{\ln(x)}{n}+o(1/n)\right)-1\right)^{n}\\ &=\left(1+\frac{\ln(x^2)}{n}+o(1/n)\right)^{n}\\ &=\exp\left(n\ln\left(1+\frac{\ln(x^2)}{n}+o(1/n)\right)\right) \end{align}$$ Can you take it from here?

Answer 2

This amounts to showing the limit of the log is equal to $2\log x$. We'll use equivalence of functions:

$$\qquad \log\left(2x^{\frac{1}{n}}-1\right)^{n}=n\log\left(2x^{\frac{1}{n}}-1\right)=\frac{\log\left(1+2(x^{\frac{1}{n}}-1)\right)}{\frac1n}\sim\frac{2(x^{\frac{1}{n}}-1)}{\frac 1n}.\qquad$$ Now $\lim_{h\to 0}\dfrac{x^h-1}h=\bigl(x^h\bigr)'_{h=0}=\bigl(x^h\log x\bigr)_{h=0}=\log x$, so the limit of the log is indeed $2\log x$.