Elegant solution to $\lim\limits_{n \to \infty}{[n(\sqrt[n]{a} - 1)]}$

Question

Exercise:

$$\lim\limits_{n \to \infty}{[n(\sqrt[n]{a} - 1)]} \text{ where a > 0}$$

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Attempt:

$\lim\limits_{n \to \infty}{[n(\sqrt[n]{a} - 1)]}$

Let $m = \sqrt[n]{a}$.

$m = \sqrt[n]{a} \longrightarrow m^n = a \longrightarrow n = \log_ma = \frac{\ln a}{\ln m}$

$\lim\limits_{n \to \infty}{\sqrt[n]{a}} = 1 \longrightarrow \lim\limits_{m \to 1}{[\frac{\ln a}{\ln m}(m - 1)]} = D$

$D = \lim\limits_{m \to 1}{[\frac{\ln a}{\ln m}(m - 1)]} = \ln a \lim\limits_{m \to 1}{\frac{m - 1}{\ln m}} = \ln a (\lim\limits_{m \to 1}{\frac{\ln m}{m - 1}})^{-1} = \ln a C^{-1} = \frac{\ln a}{ln C}$

$C = \lim\limits_{m \to 1}{\frac{\ln m}{m - 1}} = 1$ (omitted steps)

$D = \frac{\ln a}{1} = \ln a$

$$\lim\limits_{n \to \infty}{[n(\sqrt[n]{a} - 1)]} = \ln a$$

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Request:

Is there a more elegant solution to this exercise?

Answer 1

Let $n=1/x$ to get

$$\lim_{n\to\infty}n(\sqrt[n]a-1)=\lim_{x\to0^+}\frac{a^x-1}x$$

Let $a^x=u$ to get

$$\lim_{u\to1}\frac{u-1}{\log_a(u)}=\lim_{u\to1}\frac{u-1}{\ln(u)}\ln(a)=\ln(a)$$

Answer 2

Perhaps,

$$n(a^{\frac{1}{n}}-1)$$

$$=\frac{e^{\frac{1}{n} \ln a}-1}{\frac{1}{n}}$$

$$=\ln a \left(\frac{e^{\frac{1}{n} \ln a}-1}{\frac{1}{n}\ln a} \right)$$

Using,

$$\lim_{h \to 0} \frac{e^h-1}{h}=1$$

As $n \to \infty$, this approaches $(\ln a)(1)$.

Answer 3

$$ n(a^{1/n}-1) = \int_1^a x^{1/n-1} \, dx, $$ and one can show directly that $$ \int_1^a x^{1/n-1} \, dx - \log{a} = \int_1^a \left( \frac{x^{1/n}}{x} - \frac{1}{x} \right) dx = \int_1^a \left( x^{1/n} - 1 \right) \frac{dx}{x} \leqslant \left( a^{1/n} - 1 \right) \log{a} \to 0 $$ as $n \to \infty$.

Answer 4

Write $a=e^u$, so $u=\log{a}$. Then $$ n(a^{1/n}-1) = n(e^{u/n}-1) = n\left(\frac{u}{n}+O\left(\frac{1}{n^2}\right)\right) = u + O(1/n) \to u $$ as $n \to \infty$. This needs one fact about the logarithm (that it is the inverse function of the exponential), and one about the exponential: that it has a derivative on an open set containing zero.

Answer 5

This is an approach I found on MSE (will post link when I get it). The beauty of this approach is that it avoids the symbol $\log a$ completely.

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It can be directly proved via definition of Riemann integral that $$\int_{1}^{a}\frac{dx}{x}=\lim_{n\to\infty}n(\sqrt[n]{a}-1)$$ This is obvious if $a=1$ as both sides become $0$. Let $a>1$ and choose the partition $P$ of $[1,a]$ with points $x_{i} =q^{i} $ where $q^{n} =a$. And the corresponding Riemann sum is $$S(f,P)=\sum_{i=1}^{n}f(x_{i-1})(x_{i}-x_{i-1})$$ where $f(x) =1/x$. Clearly we can see that $$S(f, P) =\sum_{i=1}^{n}q^{1-i}(q^{i}-q^{i-1})= n(q-1)=n(\sqrt[n]{a}-1)$$ And since the integral $\int_{1}^{a}(dx/x)$ exists it follows that $$\lim_{n\to\infty} n(\sqrt[n]{a} - 1)=\int_{1}^{a}\frac{dx}{x}$$ If $0<a<1$ then we form the partition of interval $[a, 1]$ with point $x_{i} =aq^{i} $ where $q^{n} =1/a$ and the proof can be completed as in case of $a>1$.

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A complete theory of logarithm function can be developed either via the limit in the question or the corresponding integral which this limit equals.

Answer 6

Trivial if $a=1$. Assume then that $a\ne 1$, and let $a_n = n (a^{1/n} -1)$. Let $b _n =\frac{a_n}{n}$. Then $b_n\ne 0$, because $a^{1/n}\ne 1$ but $b_n\to 0$ because $a^{1/n}\to 1$. Now

$$a= (1+\frac{a_n}{n})^n=(1+b_n)^{\frac{1}{b_n}\times a_n}\quad\Rightarrow \quad a_n=\frac{\ln a}{\ln ( (1+b_n)^{\frac{1}{b_n}})}{\to} \frac{\ln a}{\ln e}=\ln a. $$