Let $f:\mathbb R \to \mathbb R$ given by $f\left(x\right)=\frac{a}{1+bx^2},a,b\in \mathbb R, b\ge 0$. Which of the following are true?
(A) $f$ is uniformly continuous on a compact interval of $\mathbb R$ for all values of $a,b$.
(B)$f$ is uniformly continuous on $\mathbb R$ and is bounded for for all values of $a,b$.
(C) $f$ is uniformly continuous on $\mathbb R$ only if $b=0$.
(D) $f$ is uniformly continuous on $\mathbb R$ and unbounded if $a\neq 0$ and $b\neq 0$.
My solution:-
I could prove this function is bounded for all values of $a,b$. Also know that $f$ is continuous on $\mathbb R$. I can extend the function to $\mathbb R \cup \{\infty,-\infty\}$ continuously. If it would be an interval $(a,b)$ instead of $\mathbb R$. Then I could have been saying it is uniformly continuous. Here, I am helpless. How do I prove the uniform continuity without using definition?
(B) is correct and (B) implies (A) so (A) is correct.
There exists $M\in \Bbb R^+$ such that $|f'(x)|<M$ for all $x\in \Bbb R.$
Given $\epsilon >0 ,$ if $y<z<y+\epsilon /M$ then there exists $x\in (y,z)$ such that $$|f(y)-f(z)|=|y-x|\cdot |f'(x)|<(\epsilon /M)(M)=\epsilon.$$