Limits of a sequence solved as limits of continuous function

Question

I was trying to solve the following (pointwise) limit: $$\lim _{n \rightarrow +\infty} f_n(x)=\lim _{n \rightarrow +\infty} n(x^{\frac{n+1}{n}}-x)$$ Where $\{f_n\}_{n \in \mathbb{N}}$ is the sequence of functions defined as $f_n(x)=n(x^{\frac{n+1}{n}}-x)$. And the only thing that came up in my mind was to solve it in the following way: $$\lim _{n \rightarrow +\infty} n(x^{\frac{n+1}{n}}-x)=\lim _{n \rightarrow 0} \dfrac{x^{n+1}-x}{n}=\lim_{n \rightarrow 0}\dfrac{nx\log x+o(n)}{n}=x\log x$$ Using the Maclaurin series for $g(n)=x^{n+1}-x$. Now my doubt is: considering that the Maclaurin series exist for a continuous (and perhaps differentiable several times) function, and a sequence isn't a continuous function, is it legit what I've done? More generally, can we use methods for continuous functions even for the sequences (in case the sequence is "naturally extendible" to a continuous function, like $g(n))$? If the answer is yes, why can we do this? And how can we solve this limit without using methods for continuous functions (Taylor series, De l'Hopital, and so on)?

Answer 1

For fix $x>0$ let $g(t):=x^{1+t}.$

Then

$$n(x^{\frac{n+1}{n}}-x)=\frac{g(\frac{1}{n})-g(0)}{\frac{1}{n}} \to g'(0)$$

as $n \to \infty.$