Show that $f_n(x) = \frac{\sqrt{n}x}{1+nx+n^2 x^3}$ does not converge in C([0,1]) with the $\sup$ norm

Question

Show that $\displaystyle f_n(x) = \frac{\sqrt{n}x}{1+nx+n^2 x^3}$ does not converge in C([0,1]) with the $\sup$ norm.

I first find the pointwise limit. We have $f_n(0)=0$, and for $0<x\le 1$, we have $$ 0 \le |f_n(x)| \le \frac{\sqrt{n}x}{1+nx+n^2 x^3} \le \frac{\sqrt{n}x}{n^2 x^3} = \frac{1}{n^{3/2} x^2}. $$ Taking the limit as $n \to \infty$, we see that $f_n \to f \equiv 0$ on $[0,1]$.

However, I'm not sure how to prove that this convergence is not uniform. Is there a general way to find $x=a$ in the domain $D$ so that $\sup_{x \in D} |f_n(a)-f(a)| \not\to 0$?

I tried $\lim_{n \to \infty} f_n \left( \frac{1}{\sqrt{n}} \right) = \lim_{n \to \infty} \frac{1}{1+2\sqrt{n}} = 0$.

Thank you.

Answer 1

It seems to me the CV is uniform.

Just let calculate $f'(x)=\dfrac{\sqrt{n}(1-2n^2x^3)}{D^2}$

The maximum of the function is reached for $x_m=\sqrt[3]{\frac 1{2n^2}}$

And calculation gives $f(x_m)=O(1/\sqrt{n})$ which goes to zero.