Prove that a sequence of functions is pointwise convergent

Question

I study pointwise convergence for sequences of functions. I am not sure If I do it right, so I would like to know If my way of solving it is the right way, and if it is not help me with a solution.
Here is my problem:
$f_n:[0,1] \rightarrow \mathbb{R}, f_n = nx(1-x^4)^n$ And this is my way of doing it:
First I notice $\lim_{x \rightarrow \infty}f_n(0) = \lim_{x \rightarrow \infty}f_n(1) = 0 $ What I don't know is how to prove that $\forall x \in (0,1) \lim_{x \rightarrow \infty}f_n(x) = 0$. Anyways, I assume that this sequence of functions converge pointwise to $0$.

Let $\epsilon >0 \ |f_n(x) - 0| = |f_n(x)| = |nx(1-x^4)^n| = nx(1-x^4)^n< \epsilon$. And now I should get a $N = N( \epsilon,x) \in \mathbb{N}$ but I don't know how to do it?

Is my way of solving it the right way? If so, could you please help me with the parts I am stuck with? If not, could you please help me with it?

Answer 1

You proved correctly that $\lim_{n\to\infty}f_n(1)=0$.

If $x\in[0,1)$, then$$\lim_{n\to\infty}n(1-x^4)^n=\lim_{n\to\infty}\frac n{\left(\frac1{1-x^2}\right)^n}=0$$because you have here a quotient between a polynomial function and an exponential function whose base is greater than $1$.

So,$$\lim_{n\to\infty}nx(1-x^4)^n=x\lim_{n\to\infty}n(1-x^4)^n=x\times0=0.$$