Bound of a function $f_n=\frac{x^2}{x^2+(1-nx)^2}$

Question

Let $\displaystyle f_n=\frac{x^2}{x^2+(1-nx)^2}$ where ($0\le x\le 1,\; n=1,2,3,...$)

Then $|f_n(x)| \le M$.
Find this $M$.

The answer is $1$.
Without any restriction of n, how can we find that bound?

Answer 1

For all positive integers $n$ and all $x\in[0,1]$, $0\leq x^2\leq x^2+(1-nx)^2>0$. Hence $0\leq\frac{x^2}{x^2+(1-nx)^2}\leq 1$.

Answer 2

Let $\frac{x^2}{x^2+(1-nx)^2}=y$

or $x^2(y+yn^2-1)-2nyx+y=0$

As $x$ is real, $(2ny)^2\ge 4y(y+yn^2-1)$

$\implies y^2-y\le 0\implies y(y-1)\le 0\implies 0\le y\le 1$