Why is $\lim\limits_{n\to \infty} \dfrac{x^{\frac{1}{n}}-1}{x^{\frac{s+1}{n}}-1} = \dfrac{1}{s+1}$?

Question

I believe the question tag is quite self-explanatory. Why is$$\lim_{n\to \infty} \frac{x^{\frac{1}{n}}-1}{x^{\frac{s+1}{n}}-1} = \frac{1}{s+1}$$

true? It looks like a case for l'hopitals rule, I'd say, but I'm not sure how...

Answer 1

We can use the standard limit $$\lim_{n\to \infty}n(x^{1/n} - 1)=\log x\tag{1}$$ in the following manner: \begin{align} L&= \lim_{n \to \infty}\frac{x^{1/n} - 1}{x^{(s + 1)/n} - 1}\notag\\ &= \lim_{n \to \infty}\frac{n(x^{1/n} - 1)}{n((x^{s + 1})^{1/n} - 1)}\notag\\ &= \frac{\log x}{\log x^{s + 1}}\notag\\ &= \frac{\log x}{(s + 1)\log x}\notag\\ &= \frac{1}{s + 1}\notag \end{align}

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OP wants to know more about the standard limit $(1)$ above. There are two approaches possible. One approach is to start with equation $(1)$ as the definition of $\log x$ and then develop a full theory of exponential and logarithmic functions. This is available with all the details in my blog post.

Another approach is to assume familiarity with exponential and logarithmic functions and in particular the relation $a^{b} = \exp(b\log a)$ and the limit $$\lim_{x \to 0}\frac{\exp(x) - 1}{x} = 1\tag{2}$$ Then we can see that $$\lim_{n\to\infty}n(x^{1/n} - 1) = \lim_{t \to 0}\frac{x^{t} - 1}{t} = \lim_{t \to 0}\frac{\exp(t\log x) - 1}{t\log x}\cdot\log x = \log x$$ where $t = 1/n$.

There is also another approach which shows that limit $(1)$ is equal to $\int_{1}^{x}\frac{dt}{t}$ via limit of Riemann sum.

Answer 2

we have by L'Hospital $$\lim_{n\to \infty}\frac{\frac{1}{n}x^{1/n-1}}{\frac{s+1}{n}x^{\frac{s+1}{n}-1}}$$ can you finish?