Investigate whether the sequence $\frac{nx^n + 3\sin{2n\pi x}}{n}$ is uniformly convergent Let $(f_n)$ be a sequence of functions defined by
\begin{equation*}
f_n(x) = \frac{nx^n + 3\sin{(2n\pi x)}}{n}
\end{equation*}
for all $x \in [0,1]$ and $n \in \Bbb N$.
Show that $f_n$ is converges pointwise to a function $f:[0,1]\to \Bbb R$ on $[0,1]$.
Investigate whether $f_n$ is uniformly convergent to $f$ on $[0,1]$ or not.
Attempt:

*

*For $x=0$: We have $f_n(0)=0 \to 0$.


*For $x=1$: We have $f_n(1)=1 \to 1$.


*For $x \in (0,1)$: I claimed that $f_n \to 0$. The proof is goes as follows:
Let $\varepsilon > 0$ and $x \in (0,1)$ be arbitrary. Choose $k \in \Bbb n$ such that
$k > \max\left\{\frac{2x}{\varepsilon - \varepsilon x}, \frac{6}{\varepsilon} \right\}$. Then, for any $n \in \Bbb n$ with $n \ge k$, we have
\begin{align*}
\left|f_n(x)-f(x) \right| &= \left|\frac{nx^n + 3\sin{(2n\pi x)}}{n}-0 \right| \\
&\le x^n + \frac{3}{n} \\
&= \frac{1}{\left(1+\left(\frac{1}{x}-1\right)\right)^n} + \frac{3}{n} \\
&\le \frac{1}{1+n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&< \frac{1}{n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\
&\le \frac{1}{k\left(\frac{1}{x} -1 \right)} + \frac{3}{k} \\
&< \frac{\frac{\varepsilon}{2}\left(\frac{1}{x} -1 \right)}{\frac{1}{x}-1} + \frac{\varepsilon}{2} \\
&= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align*}
Hence, $f_n \to f$ where
\begin{equation*}
f(x) =
\begin{cases}
0, \qquad \text{if} \ 0 \le x < 1 \\
1, \qquad \text{if} \ x = 1
\end{cases}.
\end{equation*}
Now, my intuition is said that $f_n$ is not uniformly convergent to $f$ on $[0,1]$. But, I didn't be able yet to find the subsequence such that the sequence did not uniformly convergent. On the other hand, I tried using the uniform norm
\begin{align*}
||f_n - f||_{[0,1]} =
    \sup \left\{\begin{array}{lr}
        \frac{nx^n + 3\sin{(2n\pi x)}}{n}, & \text{for} \ 0 \le x < 1\\
        0, & \text{for} \ x=1
        \end{array}\right\} = \frac{13}{4} \ne 0.
\end{align*}
But, is this correct? Any ideas? Thanks in advanced.

There is simple way of showing that the convergence is not uniform.
Note that $f_n(x)=x^{n}+g_n(x)$ where $g_n(x)=\frac {3\sin (2\pi nx)} n$. Now $g_n(x) \to 0$ uniformly so  $f_n$ converges uniformly if and only if $x^{n}$ does. Can you show that $x^{n}$ does not converge uniformly?
