prove equality $\lim_{n\to\infty}n(\sqrt[n]{a} - 1) = \ln a $

Question

$$\lim_{n\to\infty}n(\sqrt[n]{a} - 1) = \ln a $$ since $a > 0$

My attempt is

$\lim_{n\to\infty}n(\sqrt[n]{a} - 1) = \lim_{n\to\infty} \ln e^{n(\sqrt[n]{a} - 1)} = \lim_{n\to\infty} \ln e^{(n\sqrt[n]{a} - n)} $

$\lim_{n\to\infty} \ln e^{(n\sqrt[n]{a} - n)} = \lim_{n\to\infty} \ln \frac{e^{n\sqrt[n]{a}}}{e^n} $

And I do not know what to do then.

Answer 1

we have $\sqrt[n]{a}-1=\frac{1}{m}$ and from here we get $$n=\frac{\log(a)}{\log\left(1+\frac{1}{m}\right)}$$

Answer 2

Write the original expression as $$\frac{a^{1/n} - a^0}{\frac{1}{n} - 0}.$$ So that the limit of it can be viewed as the derivative of $a^x$ at $x = 0$.

Answer 3

Let $a= e^b$. Then you have to prove: $$ \lim_{n \to +\infty}n \left(e^{b/n}-1\right) = b. \tag{1}$$ The exponential function is a convex function, hence $e^{x}\geq x+1$ implies $n\left(e^{b/n}-1\right)\geq b$.

On the other hand, $\frac{e^x-1}{x}$ is a convex function on $(0,1]$, hence $e^{x}-1\leq x\left(1+(e-2)x\right)$ for every $x\in(0,1]$ implies that the original limit is $\color{red}{b=\log a}$ by squeezing.

Answer 4

$$\lim_{n\to\infty}n(\sqrt[n]{a} - 1) = \lim_{n\to\infty}\frac{a^{\frac{1}{n}} - 1}{\frac{1}{n}} $$ by using L'Hopital Rule $$\lim_{n\to\infty}\frac{-\frac{(\ln a)a^{\frac{1}{n}}}{n^2} }{-\frac{1}{n^2}}=\ln a$$