# Why limit of the function $\lim_{n \to \infty}{(2\sqrt[n]x-1)^n}$ where $x \ge 1$ is not $1$ but $x^2$

I was reading the below link of math.stackexchange. The question is about to solve the limit of a function Limit of a given function

The function is: $$$$f(x) = \lim_{n \to \infty}{(2\sqrt[n]x-1)^n}$$$$ where $$x \in R$$ and $$x \ge 1$$

As per my understanding,the limit of the function should be 1 but it's given $$x^2$$ in the above link.

Here is my understanding:

when $$n \to \infty$$, $$\sqrt[n]x \to 1$$ and $$(2\sqrt[n]x-1) \to 1$$.

So the limit of the funciton f(x) will also approach to $$1$$ when $$n \to \infty$$.

Can anyone explain where I am wrong.

• Your proof is not correct since it does not take into account that the exponent diverges while $(2\sqrt[n]x -1)$ converges to $1$.
– dfnu
Sep 29 '19 at 7:06

Your mistake is in thinking that if $$a_n \to 1$$ then $$a_n^{n}$$ also tends to $$1$$. This is not true. For example, $$(1+\frac 1 n)^{n} \to e$$ even though $$1 +\frac 1n \to 1$$.

• Thanks a lot to correct me. Sep 29 '19 at 7:09

You must pay attention since when $$a_n\to 1$$ we can't conclude that $$(a_n)^n\to 1$$.

Here is a sligthly different solution:

$$(2\sqrt[n]x-1)^n=(1-(2\sqrt[n]x-2))^n=e^{n\log{(1+(2-2\sqrt[n]x))}}\to x^2$$

indeed by standard limits

$$n\log{(1+(2-2\sqrt[n]x))}=\frac{\log{(1+(2-2\sqrt[n]x))}}{(2-2\sqrt[n]x))}\cdot n(2-2\sqrt[n]x-2)\to \log x^2$$

since for $$t=2-2\sqrt[n]x\to 0$$

$$\frac{\log{(1+(2-2\sqrt[n]x))}}{(2-2\sqrt[n]x))}=\log\frac{1+t}{t}\to 1$$

and

$$n(2-2\sqrt[n]x-2)=2\frac{x^\frac1n-1}{\frac1n}\to \log x^2$$