Upper bound of specific function: $\left(\frac{1}{x}-1\right)^x$

Question

Could you suggest an idea to find the upper bound of the following function:
$$\left(\frac{1}{x}-1\right)^x,$$ where $x\in (0,\frac{1}{2}]$.

Answer 1

Hint:

Use implicit differentiation to find derivative of $y=(\dfrac{1}{x}-1)^x$ and find its zero(s) and examine sign of derivative between zeros to find where the function increases and where it decreases. If it does not have zero(s), then derivative is either positive or negative on interval $(0,\dfrac{1}{2}]$ and the upper bound should be easy to find.

Answer 2

Hint: Lower Bound: $(2x)^{-x}$ and Upper Bound: $x^{-x}$ for $x\in [0,0.5]$.

We know that $x\leq \frac{1}{2}$, so, $\frac{1}{x}\geq 2$. Then, $\frac{1}{x}-1\geq 2-1=1$. Also, we know that $\frac{1}{x}\geq \frac{1}{x}-1\geq 1$. Now we just raise both sides of the inequlity to the power of $x$ we have; $(\frac{1}{x})^x\geq (\frac{1}{x}-1)^x$. Then $(\frac{1}{x})^x$ is an upper bound on [0,0.5].