Maximum of the function $f_n(x) = (1 - (1 - x^2)^n) / x$ 
Consider the following function $f_n$ defined on the interval $(0,1]$:
$$f_n(x) = \frac{1 - (1 - x^2)^n}{x}$$
Determine the maximum of this function for any natural number $n$.

I have computed the derivative of this function: $f_n(x)' = \frac{2n x^2(1 - x^2)^{n - 1} + (1 - x^2)^n - 1 }{x^2}$, and from $f_n(x)' = 0$ and by denoting $t := 1 - x^2 \in [0, 1]$, I obtained: $(1 - 2n)t^n + 2nt^{n - 1} -1= 0$.
Now to determine the roots of this polynomial in $t$ I denoted the function $g(t) = (1 - 2n)t^n + 2nt^{n - 1} -1$, and computing the derivative of $g$, we get that $g$ is increasing when $t \in [0, 1 - \frac{1}{2n - 1}]$ and is decreasing when $t \in [1 - \frac{1}{2n - 1}, 1]$. Then $g(0) = -1$, $g(1) = 0$, and $g(1 - \frac{1}{2n - 1}) > 0$. Therefore from this we deduce that the unique maximum $t$ belongs to $[0, 1 - \frac{1}{2n-1}]$, or in other words the unique $x$ that maximizes $f$ lies in $[\frac{1}{\sqrt{2n - 1}}, 1]$.
But I couldnt go beyond this point and if anyone has any ideas I would really appreciate it.
 A: I will add on to dezdichado's answer to find the exact constant on the rate of growth of the maximum. Let $x = \frac{c}{\sqrt{n}} + O\left(n^{-3/2}\right)$. Since it must be true that $$\lim_{n \to \infty}(1-2n)\left( 1-x^2 \right)^{n} + 2n\left(1-x^2 \right)^{n-1}-1=0$$
$x = \frac{c}{\sqrt{n}}$ can be plugged into this to get $$\lim_{n \to \infty}(1-2n)\left( 1-\left(\frac{c}{\sqrt{n}}\right)^2 \right)^{n} + 2n\left(1-\left(\frac{c}{\sqrt{n}}\right)^2 \right)^{n-1}-1=0$$
Equivalently, this is $$\lim_{n \to \infty}\left( 2c^2\left( 1-\frac{c^2}{n} \right)^{n-1}+\left( 1-\frac{c^2}{n} \right)^{n}-1 \right) = 0$$
Since $\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x$, this simplifies to $$(2c^2 +1)e^{-c^2}-1=0$$
$c$ is then the positive solution of $\left(2c^2+1\right)e^{-c^2}=1$. The maximal value is then given by $$\frac{2c}{2c^2+1}\sqrt{n} + O\left(n^{-1/2}\right) \approx 0.638 \sqrt{n} + O\left( n^{-1/2} \right)$$
Here are some numerical results concerning the accuracy of this approximation.
$$\left(
\begin{array}{cccc}
 n & \text{actual maximum} & \text{approximation} & \text{relative error} \\
 10 & 2.084592  &  2.018079 &    3.1907 \cdot 10^{-2} \\
 100 & 6.401867  &  6.381727 &    3.1459 \cdot 10^{-3} \\
 1000 & 20.18713  &  20.18079 &   3.1416 \cdot 10^{-4} \\
 10000 & 63.81927  &  63.81727 &    3.1411 \cdot 10^{-5} \\
 100000 & 201.8086   &  201.8079 &    3.1411 \cdot 10^{-6} \\
\end{array}
\right)$$
A: If you consider $f_n(\frac{1}{\sqrt{n}})$ for large $n$, then the numerator:
$$1 - \left(1-\frac 1n\right)^n\sim1 - e^{-1}$$
and therefore your function is at that value is roughly:
$$f_n\left(\frac{1}{\sqrt{n}}\right)\sim\sqrt{n}(1-e^{-1}).$$
This means that the maximum of $f_n(x)$ tends to infinity as $n\to\infty.$
For $n = 50,$ wolfram alpha gives the maximum as $4.54$, while the above expression is about $4.469,$ so the maximum for each given $n$ does not seem far off from the above heuristics.
Wolfram for $n = 50.$
