Uniform convergence of $f_n(x)=x^{(n-\frac{x}{n})}$ on $(0,1)$

Question

I have prove the pointwise convergence in $(0,1)$ to the null function. For the uniformly convergence I can calculate the derivative $f'_n(x)$?

If I prove $f_n$ uniformly converge, $\lim_{n\rightarrow+\infty}\int_{0}^{1} f_n(x) dx=\int_{0}^{1}\lim_{n\rightarrow+\infty} f_n(x) dx$?

$\forall n, \forall x \in (0,1) , x^{(n-\frac{x}{n})}>x^n$ so can I say Sup $x^{(n-\frac{x}{n})}$>Sup $x^n=1$ so there isn't uniformly convergence in (0,1)?

Answer 1

For any $\;x\in (0,1)\;$ we have

$$x^{n-\frac xn}=\frac{x^n}{\left(x^{1/n}\right)^x}\xrightarrow[n\to\infty]{}\frac0{1}=0$$

Answer 2

Take $$x_n = \left(1-\frac{1}{n}\right) \in (0,1)$$

Then $$f_n(x_n) \to \frac{1}{e}$$ hence $$||f_n||_\infty \ge f_n(x_n) \to \frac{1}{e}$$so the convergence is not uniform.