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Using L'hôpital rule and continuity of exponential function, we can show that

For any $a>1$

$$\lim_{x \to \infty} (2a^{\frac{1}{x}}-1)^x = a^2$$

Since the sequence is just restriction on domain to $\mathbb{N}$ of a function, the limit of the sequence is the same.

Questions is given before introducing continuity. We can use properties of real number and sequences and matric space.

niraj panakhaniya
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    Your tools seem quite limited at this point. Perhaps you might let us know what we can use. Have you tried anything yet? Have any other problems of this kind been worked in the class/book? It is hard to work this problem without using the continuity of $\log$ and $\exp$. – robjohn Aug 29 '23 at 15:23
  • Unfortunately that's the case. Tools we can use are binomial (like) expansion, lub property, Archimedian property etc. Nothing that I have tried has worked so far (used L'hôpital rule but it's not relevant). No, this was perhaps the first of a kind. It's hard so any idea or hint would be appreciated. – niraj panakhaniya Aug 29 '23 at 20:29
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    For exercises of this kind the textbook should also develop some material beforehand. In particular the book needs to discuss the convergence of $a^{1/n}$ (that's the easy part) and also the convergence of $n(a^{1/n}-1)$ (that's a bit harder). I have posted an answer based on these ideas. – Paramanand Singh Aug 30 '23 at 02:22

2 Answers2

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Let $$x_n=\frac{2a^{1/n}-1}{a^{2/n}}$$ and we can note that $$n(x_n-1)=-n(a^{1/n}-1)\cdot\frac{a^{1/n}-1} {a^{2/n}} \tag{1}$$ The first factor tends to $\log a$ (using a bit of algebra one can prove that sequence $n(a^{1/n}-1)$ is bounded and that is sufficient for the argument of this answer) and second factor tends to $0$.

Thus $n(x_n-1)\to 0 $ and hence by a well known lemma of Thomas Andrews (proved using Binomial theorem) we have $x_n^n\to 1$ ie $(2a^{1/n}-1)^n\to a^2$.

We can handle the sequence $n(a^{1/n}-1)$ (appearing as a factor in $(1)$) using Bernoulli inequality. Let $a>1$ and we set $b=a^{1/n}-1$ so that $b>0$ then $$a=(1+b) ^n\geq 1+nb$$ via Bernoulli and then $$n(a^{1/n}-1)=nb\leq a-1$$ and clearly $n(a^{1/n}-1)>0$ so that the sequence $n(a^{1/n}-1)$ is bounded.

For $0<a<1$ we can put $c=1/a$ and $$n(a^{1/n}-1)=-\frac{n(c^{1/n}-1)}{c^{1/n}}$$ The numerator is bounded and denominator tends to $1$ so that the overall fraction is bounded. It thus follows that $n(a^{1/n}-1)$ is bounded for all $a>0$.

Paramanand Singh
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Here is an argument using Bernoulli's Inequality and the Squeeze Theorem.

Elementary Approach Using Bernoulli's Inequality

For $x\ge1$, $$ \begin{align} \left(2a^{1/x}-1\right)^x &=\left(1+2\left(a^{1/x}-1\right)+\left(a^{1/x}-1\right)^2-\left(a^{1/x}-1\right)^2\right)^x\tag{1a}\\ &=\left(a^{2/x}-\left(a^{1/x}-1\right)^2\right)^x\tag{1b}\\ &=a^2\left(1-\left(1-a^{-1/x}\right)^2\right)^x\tag{1c}\\ &\ge a^2\left(1-x\left(1-a^{-1/x}\right)^2\right)\tag{1d}\\ &=a^2\left(1-x\left(1-(1+(a-1))^{-1/x}\right)^2\right)\tag{1e}\\ &\ge a^2\left(1-x\left(1-\left(1-\frac{a-1}x\right)\right)^2\right)\tag{1f}\\ &=a^2\left(1-\frac{(a-1)^2}x\right)\tag{1g} \end{align} $$ Explanation:
$\text{(1a):}$ $2a^{1/x}-1=1+2\left(a^{1/x}-1\right)$ then
$\phantom{\text{(1a):}}$ add and subtract $\left(a^{1/x}-1\right)^2$
$\text{(1b):}$ $1+2\left(a^{1/x}-1\right)+\left(a^{1/x}-1\right)^2=a^{2/x}$
$\text{(1c):}$ factor $a^2$ out front
$\text{(1d):}$ Bernoulli's Inequality (exponent $\ge1$)
$\text{(1e):}$ $a=1+(a-1)$
$\text{(1f):}$ Bernoulli's Inequality (exponent $\le0$)
$\text{(1g):}$ simplify

Using $\text{(1c)}$ and $\text{(1g)}$, we get $$ a^2\left(1-\frac{(a-1)^2}x\right)\le\left(2a^{1/x}-1\right)^x\le a^2\tag2 $$ Therefore, by the Squeeze Theorem, $$ \lim_{x\to\infty}\left(2a^{1/x}-1\right)^x=a^2\tag3 $$

Extension to Higher Powers

For $x\ge1$, $$ \begin{align} &\left(na^{1/x}-(n-1)\right)^x\\ &=\left(1+n\left(a^{1/x}-1\right)+\sum_{k=2}^n\binom{n}{k}\left(a^{1/x}-1\right)^k-\sum_{k=2}^n\binom{n}{k}\left(a^{1/x}-1\right)^k\right)^x\tag{4a}\\ &=\left(a^{n/x}-\sum_{k=2}^n\binom{n}{k}\left(a^{1/x}-1\right)^k\right)^x\tag{4b}\\ &=a^n\left(1-\sum_{k=2}^n\binom{n}{k}a^{-(n-k)/x}\left(1-a^{-1/x}\right)^k\right)^x\tag{4c}\\ &\ge a^n\left(1-x\sum_{k=2}^n\binom{n}{k}a^{-(n-k)/x}\left(1-a^{-1/x}\right)^k\right)\tag{4d}\\[3pt] &=a^n\left(1-x\left(1-a^{-n/x}-na^{-(n-1)/x}\left(1-a^{-1/x}\right)\right)\right)\tag{4e}\\[12pt] &=a^n\left(1-x\left(1-na^{-(n-1)/x}+(n-1)a^{-n/x}\right)\right)\tag{4f}\\[3pt] &=a^n\left(1-x\left(1-a^{-1/x}\right)^2\sum_{k=1}^{n-1}ka^{-(k-1)/n}\right)\tag{4g}\\ &=a^n\left(1-x\left(1-(1+(a-1))^{-1/x}\right)^2\sum_{k=1}^{n-1}ka^{-(k-1)/n}\right)\tag{4h}\\ &\ge a^n\left(1-x\left(1-\left(1-\frac{a-1}x\right)\right)^2\frac{n^2-n}2\right)\tag{4i}\\ &=a^n\left(1-\frac{(a-1)^2}x\frac{n^2-n}2\right)\tag{4j} \end{align} $$ Explanation:
$\text{(4a):}$ $na^{1/x}-(n-1)=1+n\left(a^{1/x}-1\right)$ then
$\phantom{\text{(4a):}}$ add and subtract $\sum\limits_{k=2}^n\binom{n}{k}\left(a^{1/x}-1\right)^k$
$\text{(4b):}$ $1+n\left(a^{1/x}-1\right)+\sum_{k=2}^n\binom{n}{k}\left(a^{1/x}-1\right)^k=a^{n/x}$
$\text{(4c):}$ factor $a^n$ out front
$\text{(4d):}$ Bernoulli's Inequality (exponent $\ge1$)
$\text{(4e):}$ $\sum\limits_{k=0}^n\binom{n}{k}a^{-(n-k)/x}\left(1-a^{-1/x}\right)^k=1$
$\phantom{\text{(4e):}}$ $a^{-n/x}$ is the $k=0$ term
$\phantom{\text{(4e):}}$ $na^{-(n-1)/x}\left(1-a^{-1/x}\right)$ is the $k=1$ term
$\text{(4f):}$ expand the product and collect terms
$\text{(4g):}$ $1-nx^{n-1}+(n-1)x^n=(1-x)^2\sum\limits_{k=1}^{n-1}kx^{k-1}$
$\text{(4h):}$ $a=1+(a-1)$
$\text{(4i):}$ Bernoulli's Inequality (exponent $\le0$)
$\text{(4j):}$ simplify

Using $\text{(4c)}$ and $\text{(4j)}$, we get $$ a^n\left(1-\frac{(a-1)^2}x\frac{n^2-n}2\right)\le\left(na^{1/x}-(n-1)\right)^x\le a^n\tag5 $$ Therefore, by the Squeeze Theorem, $$ \lim_{x\to\infty}\left(na^{1/x}-(n-1)\right)^x=a^n\tag6 $$

robjohn
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