Limit of the sequence $(2a^{\frac{1}{n}}-1)^n$, $a>1$. Without using L'hôpital rule or continuity.

Question

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.

Answer 1

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 $(2a^{1/n}-1)^n\to a^2$.

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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 $a>0$.

Answer 2

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

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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, $$ \lim_{x\to\infty}\left(2a^{1/x}-1\right)^x=a^2\tag3 $$

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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, $$ \lim_{x\to\infty}\left(na^{1/x}-(n-1)\right)^x=a^n\tag6 $$