Find out if a function has a maximum and minimum or not

Question

$\psi: \mathbb{R}$ → $\mathbb{R}$ is a continuous function such that $\lim_{x\to +∞} ψ(x) = +\infty $ and $\lim_{x\to -∞} ψ(x) =-\infty $

Decide if G: $\mathbb{R}$ → $\mathbb{R}$, $G(x) = \frac{ψ(x)}{1+ψ^2(x)}$ has a maximum and minimum or not.

Here's what I did: $\lim_{x\to +∞} G(x)$ = $\lim_{x\to +\infty} \frac{ψ(x)}{1+ψ^2(x)}$ = $\lim_{x\to +∞} \frac{1}{1/ψ(x)+ψ(x)}$ = $\frac{1}{+\infty} = 0$

For definition of limit, $\forall \delta \gt 0$ $\exists x \gt 0$ $\forall b \gt x $ $ |G(b)-0| \lt \delta \Leftrightarrow |G(b)| \lt \delta $

As such, in the interval $]x, +\infty[$ G has a maximum and a minimum.

In the closed and bounded interval $[0,x]$, since $G$ is continuous. Extreme value theorem says that $G$ has a maximum and a minimum.

Then I calculated the limit for $-\infty$, which was also $0$, and did the same thing for the interval $]-\infty, x[$ and $[x,0]$.

Is this correct?

Answer 1

I do have some criticisms/comments.

• Firstly, you've got a solid idea: show that the function converges to $0$ as $x \to \pm \infty$, use this fact to narrow the search down to a compact interval, and use the extreme value theorem.
• However, there's an important fact you've missed. You need to show that the function takes both positive and negative values. Otherwise, you could be getting (for example) some kind of bell-cure shape, where the maximum is achieved, but no minimum is achieved (instead, there is an infimum of $0$). If you show that the function takes both positive and negative values, then $0$ cannot be the supremum or the infimum.
• As you might expect, there are some leaps in your logic that are not adequately explained, hiding this error. The line beginning with "As such, in the interval..." is not quite justified by the previous line.
• Speaking of the previous line, you're better off fixing some $\delta > 0$. Preferably, fix it to be less than or equal to the positive value that you've guaranteed the function achieves.
• Finally (and this is the pettiest of all my criticisms), careful with the expression $\frac{1}{\infty} = 0$. You may want to cite a theorem to justify this limit. There's nothing wrong with the calculation, or the reasoning underlying it, but $\frac{1}{\infty} = 0$ is a mnemonic, not a mathematical equality.

If it were me doing this question, I would first show $\psi$ is surjective, using the definition of limits to $\infty$ and the intermediate value theorem. Let $f(x) = \frac{x}{x^2 + 1}$. I would then compute the maximum and minimum of $f$ (which is $\pm 1/2$) Then $$\max_{x \in \mathbb{R}} f(x) = \max_{x \in \mathbb{R}}f(\psi(x)) = \frac{1}{2},$$ and similarly for the minimum.

Answer 2

The function $$g(u):={u\over 1+u^2}\qquad(-\infty<u<\infty)$$ takes its minimum value $-{1\over2}$ at $u=-1$ and its maximum value ${1\over2}$ at $u=1$. Since $\psi$ is continuous it takes all values $u\in{\mathbb R}$, in particular the values $\pm1$. It follows that your $G:=g\circ\psi$ has the global minimum value $-{1\over2}$ and the global maximum value ${1\over2}$.