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: \begin{equation} f(x) = \lim_{n \to \infty}{(2\sqrt[n]x-1)^n} \end{equation} 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.

  • $\begingroup$ Your proof is not correct since it does not take into account that the exponent diverges while $(2\sqrt[n]x -1)$ converges to $1$. $\endgroup$
    – 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$.

  • $\begingroup$ Thanks a lot to correct me. $\endgroup$
    – tourism
    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 $t=2-2\sqrt[n]x\to 0$

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


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


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