Study the uniform convergence of the functional sequence $f_n(x)=\sqrt{n+1}\sin^nx\cos x$

Question

Study the uniform convergence of the functional sequence $f_n(x)=\sqrt{n+1}\sin^nx\cos x$

I found the limit

$$\lim_{n\to \infty}f_n(x)=0$$

The solution in the book it says that since

$$d_n=\sup_{x \in \mathbb R}|\sqrt{n+1}\sin^nx\cos x|=\left(\sqrt{ \frac{n}{n+1}}\right)^n \to \frac 1{\sqrt e}$$

when $n \to \infty$ then the convergence cannot be uniform in $\mathbb R$. But I don't understand how did they get that result.

Answer 1

In fact since the limit function $f(x)$ is equal to zero every where we have$$d_n=\sup_{\Bbb R} |f_n(x)-f(x)|$$The uniform convergence states that $$\forall \epsilon >0\quad,\quad\exists N\quad,\quad n>N\implies d_n<\epsilon$$which is impossible since $d_n\to {1\over \sqrt e}$ as is proved.

P.S.

$$\lim_{n\to \infty}\left(\sqrt{ \frac{n}{n+1}}\right)^n{=\exp\left(\lim_{n\to \infty}{n\over 2}\ln{n\over n+1}\right)\\=\exp\left(\lim_{u\to 0}{1\over 2u}\ln{1\over u+1}\right)\\=\exp\left(\lim_{u\to 0}-{1\over 2}{1\over u+1}\right)\\={1\over \sqrt e}}$$where we have used the L^Hopital's rule.