Calculate $\sup\limits_{x\in(0,+\infty)}\frac{x^2 e^{-n/x}}{n^2+x^2}$

Question

Calculate $$\sup_{x\in(0,+\infty)}\frac{x^2 e^{-n/x}}{n^2+x^2}$$ The derivative is $$\frac{n e^{-n/x} (n+x)^2}{(n^2+x^2)^2}\geq 0$$ then $$\sup_{x\in(0,+\infty)}\frac{x^2 e^{-n/x}}{n^2+x^2}=\lim_{x\rightarrow +\infty}\frac{x^2 e^{-n/x}}{n^2+x^2}=1$$ I would get the same result without using the derivative. We welcome suggestions.

Answer 1

For $x\gt0$, $$ \frac{x^2 e^{-n/x}}{n^2+x^2} =\frac{e^{-n/x}}{(n/x)^2+1}\tag{1} $$ Both $e^{-t}$ and $\frac1{t^2+1}$ are positive, monotonically decreasing functions for $t\ge0$. Thus, their product is also a positive, monotonically decreasing function. Therefore, $(1)$ increases as $n/x$ decreases. So, the supremum of $(1)$ would be $$ \lim_{t\to0}\frac{e^{-t}}{t^2+1}=1\tag{2} $$ since that function is continuous at $t=0$.

Answer 2

Your function can be written as

$$f(x)=\left(1-\frac{n^2}{n^2+x^2}\right)e^{-n/x}\implies f'(x)=e^{-n/x}\left(\frac{2n^2x}{(n^2+x^2)^2}+\frac n{x^2}\left(1-\frac{n^2}{n^2+x^2}\right)\right)=$$

$$=e^{-n/x}\left(\frac{2n^2x}{(n^2+x^2)^2}+\frac n{n^2+x^2}\right)\ge 0$$

and the function is monotone increasing. Also

$$\lim_{x\to 0^+}f(x)=0\cdot 0=0\;,\;\;\lim_{x\to\infty}f(x)=1\cdot e^0=1$$

and thus the wanted supremum is $\;1\;$ ( assuming $\;n\in\Bbb N\;$)