Elementary proof of $\lim_{n\to\infty}n\left(a^{\frac{1}{n}}-1\right)=\ln a$ I'm trying to proof that
$$\lim_{n\to\infty}n\left(a^{\frac{1}{n}}-1\right)=\ln a$$
without using limit of function. I can use only sequence limits. The most elementary proof.
Below you can find my solution, but I'm not sure if it's OK.
Started with limit of
$$\lim_{n\to\infty}\frac{\frac{1}{n}}{\log_a\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty}\frac{1}{n\log_a\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty}\frac{1}{\log_a\left(1+\frac{1}{n}\right)^n}=\frac{1}{\log_a e}=\ln a$$
Then let's create new sequence $b_n$ such that $b_n=a^{\frac{1}{n}}-1$. From definition of $b_n$ I can find expresion for $\frac{1}{n}$
$$
b_n=a^{\frac{1}{n}}-1\Rightarrow b_n+1=a^{\frac{1}{n}}\Rightarrow \log_a\left(b_n+1\right)=\frac{1}{n}
$$
We can see that
$$
\lim_{n\to\infty}b_n=0=\lim_{n\to\infty}\frac{1}{n}
$$
Next step is problematic for me. $\lim_{n\to\infty}\frac{1}{n}=0$ and $\lim_{n\to\infty}b_n=0$. So I can replace those expressions (can I?)
Then
$$
\lim_{n\to\infty}n\left(a^{\frac{1}{n}}-1\right)=\lim_{n\to\infty}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}\overbrace{=}^{??}\lim_{n\to\infty}\frac{b_n}{\log_a\left(b_n+1\right)}=\ln a
$$
Is there any theorem about replacing similar sequences to calculate limits? Without involving definition of limit of function. Is there any simpler proof (from some inequalities for instance) of this limit?
 A: Consider letting $a = e^b$ such that the expression becomes
$$\lim_{n\to\infty}n\left(e^{\frac{b}{n}}-1\right) = \lim_{n\to\infty}n\left((e^b)^{\frac{1}{n}}-1\right)$$
Since
$$e^b = \lim_{n \rightarrow \infty} \Big(1 + \frac{b}{n}\Big)^n (*)$$
The expression becomes
$$\lim_{n\to\infty}n\left(1+ \frac{b}{n}-1\right) = b = \log(a)$$
(Note: $(*)$ can be proven easily if you let $e^b = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^{bn}$, and let $z = bn$ such that $z$ also goes to infinity, so $n = \frac{z}{b}$ and the expression can be re-written as $e^b = \lim_{z \rightarrow \infty} (1 + \frac{b}{z})^{z}.$ This works assuming $b$ is non-negative, and expanding it to include negatives is not hard.
A: Remark that $\lim\limits_{X\to 0}\frac{e^X-1}{X} = 1$
$$\lim_{n\to\infty}n\left(a^{\frac{1}{n}}-1\right) = \lim_{n\to\infty}\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}} = \lim_{n\to\infty}\frac{e^{\frac{1}{n}\ln a}-1}{\frac{1}{n}} = \lim_{n\to\infty}\frac{e^{\frac{1}{n}\ln a}-1}{\frac{1}{n}\ln a}\ln a = \lim_{X\to0}\frac{e^{X}-1}{X}\ln a = \ln a$$
A: It's difficult to predict whether it will seem simple or not, but I'll be based on simple inequalities and fact $e^{\frac{1}{n}}\to 1$:
knowing $\left(1+\frac{1}{n} \right)^n <e < \left(1+\frac{1}{n-1} \right)^n$, we can found $1 < n\left(e^{\frac{1}{n}} -1\right) < 1+ \frac{1}{n-1}$ i.e. $\lim\limits_{n \to \infty}n\left(e^{\frac{1}{n}} -1\right) = 1$.
Now let's consider $a>1$. Then  $y_n=n\left(a^{\frac{1}{n}} -1\right) = n \left(e^{\frac{\ln a}{n}} -1\right) = z_n\left(e^{\frac{1}{z_n}} -1\right)\ln a$. Where $z_n=\frac{n}{\ln a}\to +\infty$. Let's denote $\alpha_n=\left\lfloor z_n \right\rfloor$, then we have estimations:
$$\ln a \cdot \alpha_n \left(e^{\frac{1}{\alpha_n+1}} -1\right) < y_n < \ln a \cdot (\alpha_n +1) \left(e^{\frac{1}{\alpha_n}} -1\right)\quad(1)$$
Now, because subsequence have same limit as sequence, holds $\lim\limits_{n \to \infty}\alpha_n \left(e^{\frac{1}{\alpha_n}} -1\right)=1$ and left and right sides in $(1)$ will have limit $\ln a$ i.e. it will have $y_n$.
Case $0<a<1$ comes from considered.
