Limit representation for $\log(x)$

Question

I know you can express $\log(x)$ as $$ \lim_ {n\rightarrow\infty} n (x^{1/n} - 1) $$

But I'm have a hard time getting started. Any hints?

Answer 1

We can write the function as $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}\right)\;$

Now put $\frac{1}{n} = p\;,$ so when $\lim_{n\rightarrow \infty}\;$ then $p\rightarrow 0$

So the limit converts itself into $\displaystyle \lim_{p\rightarrow 0}\left(\frac{x^p-1}{p}\right)$

Now using $\bf{L'Hôpital\; Rule}$

We get $\displaystyle \lim_{p\rightarrow 0}\left(\frac{x^{p}\ln x-0}{1}\right) = \ln x$

Answer 2

Since $x>0$, $x=e^{t}$ for some $t\in\mathbb{R}$. Then: $$ \lim_{n\to +\infty} n \left(x^{1/n}-1\right) = \lim_{n\to +\infty}n\left(e^{\frac{t}{n}}-1\right)=\lim_{z\to 0}\frac{e^{tz}-1}{z}=t=\log(x),$$ by the well-known fact: $$ \lim_{z\to 0}\frac{e^z-1}{z}=1.$$

Answer 3

Consider $$ \int_1^x t^{1/n-1} \,dt = \frac{x^{1/n}-1}{1/n}, $$ and consider taking the limit as $n \to 0$. It is easy to show that the LHS is an decreasing function of $n$ with greatest lower bound $\int_1^x t^{-1}\, dx = \log{x}$ (using that $t^{1/n}>1$, and so on).

Answer 4

Using $\int x^m \,dx = \frac{1}{m+1}x^{m+1} + C$, we have

$$\ln|x| = \lim_{n\rightarrow\infty} \int x^{1/n-1} \,dx = \lim_{n\rightarrow\infty} nx^{1/n} + K = \lim_{n\rightarrow\infty} nx^{1/n} - n = \lim_{n\rightarrow\infty} n(x^{1/n} - 1)$$

as $\ln1 = 0 \Rightarrow K = -n.$

See also this question.

Answer 5

Invoke the substitution $n=\frac1t$. Then we could turn the limit to $$\lim_{\frac1t\rightarrow\infty}\frac{x^t-1}t=\lim_{t\rightarrow0}\frac{x^t-1}t$$Which is the definition of the derivative of $x^n$ with respect to $n$ at $n=0$.